Mass of a negatively charged muon, mμ = 207me According to Bohr's model, Bohr radius, re ∝ 1/ me And, energy of a ground state electronic hydrogen atom, Ee ∝ me. Also, the energy of a ground state muonic hydrogen atom, Eu ∝ mu. We have the value of the first Bohr orbit, re = 0.53 A = 0.53 x 10⁻¹⁰m LRead more

Mass of a negatively charged muon, mμ = 207m_{e}

According to Bohr’s model,

Bohr radius, re ∝ 1/ m_{e}

And, energy of a ground state electronic hydrogen atom, E_{e} ∝ m_{e}.

Also, the energy of a ground state muonic hydrogen atom, E_{u} ∝ m_{u}.

We have the value of the first Bohr orbit, r_{e} = 0.53 A = 0.53 x 10⁻¹⁰m

Let rμ,j be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

mμ rμ = m_{e }r_{e}

207 m_{e x }rμ = m_{e }r_{e}

Therefore, rμ = 0.53 x 10⁻¹⁰ /207 = 2.56 x 10⁻¹³ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 x 10⁻^{13} m.

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 x 10⁻^{13} m

We have ,

Ee = -13.6 eV

Take the ratio of these energies as:

Ee/Eμ = me/mμ = me/207me

Eμ = 207 Ee = 207x(-13.6) = -2.8l keV
Hence, the ground state energy of a muonic hydrogen atom is -2.81 keV

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck's constant (h). The angular momentum of the Earth in its orbit is of the order of 1070h. This leads to a very high value of quanRead more

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 10^{70}h. This leads to a very high value of quantum levels n of the order of 10^{70}. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

Ans (a). Total energy of the electron, E = -3.4 eV Kinetic energy of the electron is equal to the negative of the total energy. => K = -E => - (- 3.4) = +3.4 eV Hence, the kinetic energy of the electron in the given state is +3.4 eV. Ans (b). Potential energy (U) of the electroRead more

Ans (a).

Total energy of the electron, E = -3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy.

=> K = -E => – (- 3.4) = +3.4 eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

Ans (b).

Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.

=» U = -2 K => – 2 x 3.4 = – 6.8 eV

Hence, the potential energy of the electron in the given state is – 6.8 eV.

Ans (c).
The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

Ans (a). Charge on an electron, e = 1.6 x 10⁻¹⁹C Mass of an electron, me = 9.1 x 10⁻³¹ kg Speed of light, c = 3 x 10⁸ m/s Let us take a quantity involving the given quantities as [e²/[(4π ε0mec² )] Where, ε0 = Permittivity of free space and 1/4π ε0 = 9 x 10⁹ Nm²C⁻² The numerical value of the takenRead more

Ans (a).

Charge on an electron, e = 1.6 x 10⁻¹⁹C

Mass of an electron, m_{e} = 9.1 x 10⁻³¹ kg

Speed of light, c = 3 x 10⁸ m/s

Let us take a quantity involving the given quantities as [e²/[(4π ε0mec² )]

Where,

ε0 = Permittivity of free space and

1/4π ε0 = 9 x 10⁹ Nm²C⁻²
The numerical value of the taken quantity will be:

1/4π ε0 x e²/m_{e}c² = 9 x 10⁹ x (1.6 x 10⁻¹⁹)² /( 9.1 x 10⁻³¹) x ( 3 x 10⁸)²

= 2.81 x 10⁻¹⁵ m

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

Ans (b).

Charge on an electron, e = 1.6 x 10⁻¹⁹ C

Mass of an electron, m_{e} =9.1 x 10⁻³¹ kg

Planck’s constant, h = 6.63 x 10⁻³⁴ Js

Let us take a quantity involving tile given quantities as

4π ε0 (h/2π)² /m_{e} e²

Where,

ε0 = Permittivity of free space and

1/4π ε0 = 9 x 10⁹ Nm²C⁻²

The numerical value of the taken quantity will be: 1/4π ε0 x (h/2π)²/m_{e} e² =

= 1/(9 x 10⁹) x [(6.63 x 10⁻³⁴ ) /(2 x 3.14)]²/( 9.1 x 10⁻³¹)(1.6 x 10⁻¹⁹)²

= 0.53 x 10⁻¹⁰ m

Hence, the value of the quantity taken is of the order of the atomic size.

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1]. We have the relation for energy (E1) of radiation at level n as: E₁ = hν₁ = hme⁴ /[(4π )³ ε0²(h/2π )³] x (1/n²)---------------- Eq-1 Where, ν₁= Frequency of radiation at level n h= Planck's constantRead more

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1].

We have the relation for energy (E1) of radiation at level n as:

Radius of the first Bohr orbit is given by the relation, r1 = 4πε0 (h/2π)²/mee² ------------- Eq-1 Where, Eo = Permittivity of free space h = Planck's constant = 6.63 x 10-34 Js me= Mass of an electron = 9.1 x 10-³¹kg e = Charge of an electron = 1.9 x 10-19 C mp= Mass of a proton = 1.67 x 10 -27 kgRead more

Radius of the first Bohr orbit is given by the relation,

r1 = 4πε0 (h/2π)²/m_{e}e² ————- Eq-1

Where,

Eo = Permittivity of free space

h = Planck’s constant = 6.63 x 10^{-34} Js

m_{e}= Mass of an electron = 9.1 x 10^{-³¹}kg

e = Charge of an electron = 1.9 x 10^{–}^{19} C

m_{p}= Mass of a proton = 1.67 x 10 ^{–}^{27} kg

r= Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as:

F_{C} = e²/4πε0r²————-Eq-2

Gravitational force of attraction between an electron and a proton is given as:

F_{G} =G m_{p} m_{e} /r² ——————- Eq-3 ,

^{F}c, =-^H- -(3)

Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m^{2}/kg^{2}

If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

Therefore , F_{G}=F_{C}

=> G m_{p} m_{e} /r² = e²/4πε0r²

Therefore ,

e²/4πε0 =G m_{p} m_{e}—————Eq-4

Putting the value of equation (4) in equation (1),we get:

r1= (h/2π)²/G m_{p} m²_{e}

= [6.63 x 10^{-34}/(2 x 3.14)]²/(6.67 x 10⁻¹¹) x (1.67 x 10 ^{–}^{27}) x (9.1 x 10^{-³¹})²

≈1.21 x 10²⁹
It is known that the universe is 156 billion light years wide or 1.5 x 10^{27} m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.

Ans (a). about the same The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models. Ans (b).much less The probability of scattering of a-particleRead more

Ans (a).

about the same

The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

Ans (b).much less

The probability of scattering of a-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

Ans (c).

Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

Ans (d).

Thomson’s model

It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of a-particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m Orbital speed of the Earth, v = 3 x 104 m/s Mass of the Earth, m = 6.0 x 10²4 kg According to Bohr's model, angular momentum is quantized and given as: mvr = nh/2π Where, h = Planck's constant = 6.62 x 10⁻34 Js n = Quantum number TherefRead more

Rdius of the orbit of the Earth around the Sun, r= 1.5 x 10^{11} m

Orbital speed of the Earth, v = 3 x 10^{4} m/s

Mass of the Earth, m = 6.0 x 10²^{4} kg

According to Bohr’s model, angular momentum is quantized and given as:

mvr = nh/2π

Where,

h = Planck’s constant = 6.62 x 10⁻^{34} Js

n = Quantum number

Therefore , n = mvr2π/h

= 2π x (6.0 x 10²^{4} ) x (3 x 10^{4} ) x (1.5 x 10^{11}) /(6.62 x 10⁻^{34})

= 25.61 x 10⁷³= 2.6 x 10⁷⁴

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 10^{74}.

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogenRead more

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e.,

-1.1 eV.

Orbital energy is related to orbit level (n) as:

E = -13.6/(n)² eV

For n=3 ,

E = -13.6/9 = -1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

1/λ =Ry (1/1² – 1/n²)

Where,

R_{y} = Rydberg constant = 1.097 x 10^{7} m⁻^{1}

λ= Wavelength of radiation emitted by the transition of tile electron. For n = 3, we can obtain λ as:

1/λ = 1.097 x 10^{7} (1/1² – 1/3²)

= 1.097 x 10^{7 }(1-1/9) = 1.097 x 10^{7 }(8/9)

λ =9/8 x ( 1.097 x 10^{7 }) = 102.55 nm

if the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

1/λ = 1.097 x 10^{7} (1/1² – 1/2²)

= 1.097 x 10^{7 }(1-1/4) = 1.097 x 10^{7 }(3/4)

λ = 4/( 1.097 x 10^{7 })x(3) = 121.54 nm If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:
1/λ =1.097 x 10^{7 }(1/2²-1/3²)

= 1.097 x 10^{7 }(1/4 -1/9) =(1.097 x 10^{7 }) x 5/36

λ= 36/ (5 x 1.097 x 10^{7}) = 656.33 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence ,in Lyman series ,two wavelengths i.e. 102.5 nm and 121.5nm are emitted .And in the Balmer series, one wavelength i.e.,656.33nm is emitted.

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m. Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as: r2 = (n)²r1 = 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m For n = 3, we can write the corresponding electron radius as: r3 = (n)²r1 = 9 xRead more

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻^{11} m.

Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:

r2 = (n)²r1

= 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m

For n = 3, we can write the corresponding electron radius as:

r3 = (n)²r1

= 9 x 5.3 x 10⁻¹¹=4.77 x 10⁻^{10} m

Hence, the radii of an electron for n = 2 and n = 3 orbits are 22.12 x 10⁻¹⁰ m and 4.77 x 10⁻^{10} m respectively.

## Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom [i.e., an atom in which a negatively charged muon (μ– ) of mass about 207me orbits around a proton].

Mass of a negatively charged muon, mμ = 207me According to Bohr's model, Bohr radius, re ∝ 1/ me And, energy of a ground state electronic hydrogen atom, Ee ∝ me. Also, the energy of a ground state muonic hydrogen atom, Eu ∝ mu. We have the value of the first Bohr orbit, re = 0.53 A = 0.53 x 10⁻¹⁰m LRead more

Mass of a negatively charged muon, mμ = 207m

_{e}According to Bohr’s model,

Bohr radius, re ∝ 1/ m

_{e}And, energy of a ground state electronic hydrogen atom, E

_{e}∝ m_{e}.Also, the energy of a ground state muonic hydrogen atom, E

_{u}∝ m_{u}.We have the value of the first Bohr orbit, r

_{e}= 0.53 A = 0.53 x 10⁻¹⁰mLet rμ,j be the radius of muonic hydrogen atom.

At equilibrium, we can write the relation as:

mμ rμ = m

_{e }r_{e}207 m

_{e x }rμ = m_{e }r_{e}Therefore, rμ = 0.53 x 10⁻¹⁰ /207 = 2.56 x 10⁻¹³ m

Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 x 10⁻

^{13}m.Hence, the value of the first Bohr radius of a muonic hydrogen atom is 2.56 x 10⁻

^{13}mWe have ,

Ee = -13.6 eV

Take the ratio of these energies as:

Ee/Eμ = me/mμ = me/207me

Eμ = 207 Ee

See less= 207x(-13.6) = -2.8l keV

Hence, the ground state energy of a muonic hydrogen atom is -2.81 keV

## If Bohr’s quantisation postulate (angular momentum = nh/2π) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun?

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck's constant (h). The angular momentum of the Earth in its orbit is of the order of 1070h. This leads to a very high value of quanRead more

We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Planck’s constant (h). The angular momentum of the Earth in its orbit is of the order of 10

See less^{70}h. This leads to a very high value of quantum levels n of the order of 10^{70}. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.## The total energy of an electron in the first excited state of the hydrogen atom is about –3.4 eV. (a) What is the kinetic energy of the electron in this state? (b) What is the potential energy of the electron in this state? (c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Ans (a). Total energy of the electron, E = -3.4 eV Kinetic energy of the electron is equal to the negative of the total energy. => K = -E => - (- 3.4) = +3.4 eV Hence, the kinetic energy of the electron in the given state is +3.4 eV. Ans (b). Potential energy (U) of the electroRead more

Ans (a).Total energy of the electron, E = -3.4 eV

Kinetic energy of the electron is equal to the negative of the total energy.

=> K = -E => – (- 3.4) = +3.4 eV

Hence, the kinetic energy of the electron in the given state is +3.4 eV.

Ans (b).Potential energy (U) of the electron is equal to the negative of twice of its kinetic energy.

=» U = -2 K => – 2 x 3.4 = – 6.8 eV

Hence, the potential energy of the electron in the given state is – 6.8 eV.

See lessAns (c).The potential energy of a system depends on the reference point taken. Here, the potential energy of the reference point is taken as zero. If the reference point is changed, then the value of the potential energy of the system also changes. Since total energy is the sum of kinetic and potential energies, total energy of the system will also change.

## Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10⁻¹⁰m). (a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value. (b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, me, and e will yield the right atomic size. Construct a quantity with the dimension of length from h, me, and e and confirm that its numerical value has indeed the correct order of magnitude.

Ans (a). Charge on an electron, e = 1.6 x 10⁻¹⁹C Mass of an electron, me = 9.1 x 10⁻³¹ kg Speed of light, c = 3 x 10⁸ m/s Let us take a quantity involving the given quantities as [e²/[(4π ε0mec² )] Where, ε0 = Permittivity of free space and 1/4π ε0 = 9 x 10⁹ Nm²C⁻² The numerical value of the takenRead more

Ans (a).Charge on an electron, e = 1.6 x 10⁻¹⁹C

Mass of an electron, m

_{e}= 9.1 x 10⁻³¹ kgSpeed of light, c = 3 x 10⁸ m/s

Let us take a quantity involving the given quantities as [e²/[(4π ε0mec² )]

Where,

ε0 = Permittivity of free space and

1/4π ε0 = 9 x 10⁹ Nm²C⁻²

The numerical value of the taken quantity will be:

1/4π ε0 x e²/m

_{e}c² = 9 x 10⁹ x (1.6 x 10⁻¹⁹)² /( 9.1 x 10⁻³¹) x ( 3 x 10⁸)²= 2.81 x 10⁻¹⁵ m

Hence, the numerical value of the taken quantity is much smaller than the typical size of an atom.

Ans (b).Charge on an electron, e = 1.6 x 10⁻¹⁹ C

Mass of an electron, m

_{e}=9.1 x 10⁻³¹ kgPlanck’s constant, h = 6.63 x 10⁻³⁴ Js

Let us take a quantity involving tile given quantities as

4π ε0 (h/2π)² /m

_{e}e²Where,

ε0 = Permittivity of free space and

1/4π ε0 = 9 x 10⁹ Nm²C⁻²

The numerical value of the taken quantity will be: 1/4π ε0 x (h/2π)²/m

_{e}e² == 1/(9 x 10⁹) x [(6.63 x 10⁻³⁴ ) /(2 x 3.14)]²/( 9.1 x 10⁻³¹)(1.6 x 10⁻¹⁹)²

= 0.53 x 10⁻¹⁰ m

Hence, the value of the quantity taken is of the order of the atomic size.

See less## Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n–1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1]. We have the relation for energy (E1) of radiation at level n as: E₁ = hν₁ = hme⁴ /[(4π )³ ε0²(h/2π )³] x (1/n²)---------------- Eq-1 Where, ν₁= Frequency of radiation at level n h= Planck's constantRead more

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n-1].

We have the relation for energy (E1) of radiation at level n as:

E₁ = hν₁ = hme⁴ /[(4π )³ ε0²(h/2π )³] x (1/n²)—————- Eq-1

Where,

ν₁= Frequency of radiation at level n

h= Planck’s constant

m = Mass of hydrogen atom

e = Charge on an electron

ε0= Permittivity of free space

Now, the relation for energy (Ez) of radiation at level (n – 1) is given as:

E2 = hν2 = hme⁴ /[(4π )³ ε0²(h/2π )³] x 1/[(n-1)²]————Eq-2

Where.

ν2 = Frequency of radiation at level (n-1)

Energy (E) released as a result of de-excitation:

E – E2-E1

hν = E2-E1 —————–Eq-3

Where,

ν = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii], we get:

ν =me⁴ /[(4π )³ ε0²(h/2π )³] x{1/(n-1)² – 1/n² }

= me⁴ (2n-1)/[(4π )³ ε0²(h/2π )³] n² (n-1)²

For large n, we can write (2n-1) ≈ 2n and (n-1) ≈n.

Therefore, ν = me⁴/32 π³ ε0² (h/2π )³ n³—————-Eq-4

Classical relation of frequency of revolution of an electron is given as:

νc = ν/2πr,—————Eq-5

Where,

Velocity of the electron in the nth orbit is given as :

v = e²/[(4π ε0)(h/2π )ⁿ]—————-Eq-6

And radius of nth orbit is given as :

r= [4π ε0 (h/2π )²] n²/me² —————-Eq-7

Putting the values of Eq-6 and Eq-7 in Eq-5 we get

vc = me⁴/32 π³ ε0² (h/2π )³ n³—————-Eq-8

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

See less## The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10⁻⁴⁰ .An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Radius of the first Bohr orbit is given by the relation, r1 = 4πε0 (h/2π)²/mee² ------------- Eq-1 Where, Eo = Permittivity of free space h = Planck's constant = 6.63 x 10-34 Js me= Mass of an electron = 9.1 x 10-³¹kg e = Charge of an electron = 1.9 x 10-19 C mp= Mass of a proton = 1.67 x 10 -27 kgRead more

Radius of the first Bohr orbit is given by the relation,

r1 = 4πε0 (h/2π)²/m

_{e}e² ————- Eq-1Where,

Eo = Permittivity of free space

h = Planck’s constant = 6.63 x 10

^{-34}Jsm

_{e}= Mass of an electron = 9.1 x 10^{-³¹}kge = Charge of an electron = 1.9 x 10

^{–}^{19}Cm

_{p}= Mass of a proton = 1.67 x 10^{–}^{27}kgr= Distance between the electron and the proton

Coulomb attraction between an electron and a proton is given as:

F

_{C}= e²/4πε0r²————-Eq-2Gravitational force of attraction between an electron and a proton is given as:

F

_{G}=G m_{p}m_{e}/r² ——————- Eq-3 ,^{F}c, =-^H- -(3)Where, G = Gravitational constant = 6.67 x 10⁻¹¹ N m

^{2}/kg^{2}If the electrostatic (Coulomb) force and the gravitational force between an electron and a proton are equal, then we can write:

Therefore , F

_{G}=F_{C}=> G m

_{p}m_{e}/r² = e²/4πε0r²Therefore ,

e²/4πε0 =G m

_{p}m_{e}—————Eq-4Putting the value of equation (4) in equation (1),we get:

r1= (h/2π)²/G m

_{p}m²_{e}= [6.63 x 10

^{-34}/(2 x 3.14)]²/(6.67 x 10⁻¹¹) x (1.67 x 10^{–}^{27}) x (9.1 x 10^{-³¹})²≈1.21 x 10²⁹

See lessIt is known that the universe is 156 billion light years wide or 1.5 x 10

^{27}m wide. Hence, we can conclude that the radius of the first Bohr orbit is much greater than the estimated size of the whole universe.## Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better. (a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (b) Is the probability of backward scattering (i.e., scattering of α-particles at angles greater than 90°) predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model? (c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide? (d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α-particles by a thin foil?

Ans (a). about the same The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford's model. This is because the average angle was taken in both models. Ans (b).much less The probability of scattering of a-particleRead more

Ans (a).about the same

The average angle of deflection of a-particles by a thin gold foil predicted by Thomson’s model is about the same size as predicted by Rutherford’s model. This is because the average angle was taken in both models.

Ans (b).much lessThe probability of scattering of a-particles at angles greater than 90° predicted by Thomson’s model is much less than that predicted by Rutherford’s model.

Ans (c).Scattering is mainly due to single collisions. The chances of a single collision increases linearly with the number of target atoms. Since the number of target atoms increase with an increase in thickness, the collision probability depends linearly on the thickness of the target.

Ans (d).Thomson’s model

It is wrong to ignore multiple scattering in Thomson’s model for the calculation of average angle of scattering of a-particles by a thin foil. This is because a single collision causes very little deflection in this model. Hence, the observed average scattering angle can be explained only by considering multiple scattering.

See less## In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)

Rdius of the orbit of the Earth around the Sun, r= 1.5 x 1011 m Orbital speed of the Earth, v = 3 x 104 m/s Mass of the Earth, m = 6.0 x 10²4 kg According to Bohr's model, angular momentum is quantized and given as: mvr = nh/2π Where, h = Planck's constant = 6.62 x 10⁻34 Js n = Quantum number TherefRead more

Rdius of the orbit of the Earth around the Sun, r= 1.5 x 10

^{11}mOrbital speed of the Earth, v = 3 x 10

^{4}m/sMass of the Earth, m = 6.0 x 10²

^{4}kgAccording to Bohr’s model, angular momentum is quantized and given as:

mvr = nh/2π

Where,

h = Planck’s constant = 6.62 x 10⁻

^{34}Jsn = Quantum number

Therefore , n = mvr2π/h

= 2π x (6.0 x 10²

^{4}) x (3 x 10^{4}) x (1.5 x 10^{11}) /(6.62 x 10⁻^{34})= 25.61 x 10⁷³= 2.6 x 10⁷⁴

Hence, the quanta number that characterizes the Earth’ revolution is 2.6 x 10

See less^{74}.## A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV. When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogenRead more

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes -13.6 + 12.5 eV i.e.,

-1.1 eV.

Orbital energy is related to orbit level (n) as:

E = -13.6/(n)² eV

For n=3 ,

E = -13.6/9 = -1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

1/λ =Ry (1/1² – 1/n²)

Where,

R

_{y}= Rydberg constant = 1.097 x 10^{7}m⁻^{1}λ= Wavelength of radiation emitted by the transition of tile electron. For n = 3, we can obtain λ as:

1/λ = 1.097 x 10

^{7}(1/1² – 1/3²)= 1.097 x 10

^{7 }(1-1/9) = 1.097 x 10^{7 }(8/9)λ =9/8 x ( 1.097 x 10

^{7 }) = 102.55 nmif the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

1/λ = 1.097 x 10

^{7}(1/1² – 1/2²)= 1.097 x 10

^{7 }(1-1/4) = 1.097 x 10^{7 }(3/4)λ = 4/( 1.097 x 10

^{7 })x(3) = 121.54 nmIf the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

1/λ =1.097 x 10

^{7 }(1/2²-1/3²)= 1.097 x 10

^{7 }(1/4 -1/9) =(1.097 x 10^{7 }) x 5/36λ= 36/ (5 x 1.097 x 10

^{7}) = 656.33 nmThis radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence ,in Lyman series ,two wavelengths i.e. 102.5 nm and 121.5nm are emitted .And in the Balmer series, one wavelength i.e.,656.33nm is emitted.

See less## The radius of the innermost electron orbit of a hydrogen atom is 5.3×10⁻⁻¹¹ m. What are the radii of the n = 2 and n =3 orbits?

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻11 m. Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as: r2 = (n)²r1 = 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m For n = 3, we can write the corresponding electron radius as: r3 = (n)²r1 = 9 xRead more

The radius of the innermost orbit of a hydrogen atom, r1 = 5.3 x 10⁻

^{11}m.Let r2 be the radius of the orbit at n = 2. It is related to the radius of the innermost orbit as:

r2 = (n)²r1

= 4 x 5.3 x 10⁻¹¹ = 2.12 x 10⁻¹⁰ m

For n = 3, we can write the corresponding electron radius as:

r3 = (n)²r1

= 9 x 5.3 x 10⁻¹¹=4.77 x 10⁻

^{10}mHence, the radii of an electron for n = 2 and n = 3 orbits are 22.12 x 10⁻¹⁰ m and 4.77 x 10⁻

See less^{10}m respectively.