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  1. Asked: In:

    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].

    Best Answer
    Pawan1308

    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 = 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

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

    At equilibrium, we can write the relation as:

    mμ rμ = me  re

    207 me x rμ = me  re

    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

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  2. Asked: In:

    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?

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    Pawan1308

    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 1070h. This leads to a very high value of quantum levels n of the order of 1070. For large values of n, successive energies and angular momenta are relatively very small. Hence, the quantum levels for planetary motion are considered continuous.

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  3. Asked: In:

    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?

    Best Answer
    Pawan1308

    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.

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  4. Asked: In:

    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.

    Best Answer
    Pawan1308

    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, 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 taken quantity will be:

    1/4π ε0  x   e²/mec² = 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, me =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 /me 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/me 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.

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  6. Asked: In:

    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.

    Best Answer
    Pawan1308

      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.

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