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

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.

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.

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.

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 10⁷ 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 10⁷ (1/1²   –  1/3²)

= 1.097 x 10⁷ (1-1/9) = 1.097 x 10⁷ (8/9)

λ =9/8 x ( 1.097 x 10⁷ ) = 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 10⁷ (1/1²   –  1/2²)

= 1.097 x 10⁷ (1-1/4) = 1.097 x 10⁷ (3/4)

λ = 4/( 1.097 x 10⁷ )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 10⁷  (1/2²-1/3²)

= 1.097 x 10⁷  (1/4 -1/9) =(1.097 x 10⁷ ) x 5/36

λ= 36/ (5 x 1.097 x 10⁷) = 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.