Atoms and Nuclei: Important Questions

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Very Short Answer Question [1 Mark Questions]

Ques. Name the hydrogen spectrum series lying in ultraviolet and visible regions.

Ans.  In Ultraviolet Region: Lyman series. And in the visible region: Balmer series

Ques. What is Bohr’s quantization condition for the angular momentum of an electron in the second orbit?

Ans.  

Ques. The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What is the radius of the orbit in the second excited state?

Ans. Radius (r) = n² × 5.3 × 10^-11 m

The radius of the second excited state (n =3) is:

r = (3)² × 5.3 × 10^-11 m

= 9 × 5.3 × 10^-11 m

= 4.77 × 10^-10 m

Ques. Define ionization energy.

Ans. Ionization energy: The energy needed to bash out an electron from an atom is known as the atom's ionization energy.

Ques. The base state energy of a hydrogen atom is -13.6 eV. What are the kinetic and potential energies of electrons in this condition?

Ans. Kinetic energy (Ke) = + T.E. = 13.6 eV

Potential energy (Pe) = 2 T.E. 

= 2 (-13.6)

= – 27.2 eV

Ques. What is the Q-value of a nuclear reaction?

Ans. Q−value = Mass of reactants−Mass of products

Ques. Write the empirical relation for Paschen series lines of hydrogen atoms.

Ans. 1/λ=R(1/3²−1/n²)

where n=4,5,6,7......

Ques. Complete the following nuclear reactions:

a) ₄Be⁹ + ₁H¹ → ₃Li⁶ + .......... 

b) ₅B¹⁰ + ₂He⁴ → ₇N¹³ + .......... 

Ans. a): ₄Be⁹ + ₁H¹ → ₃Li⁶ + ₂He⁴

b): ₅B¹⁰ + ₂He⁴ → ₇N¹³ + ₀n¹ 

Also check: Potential Energy

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Short Answer Question [2 Marks Questions]

Ques. Define Bohr’s radius.

Ans. Bohr's Radius: The radius (r) of the first orbit of a hydrogen atom is termed Bohr's radius. The value of Bohr's radius is 5.29×10⁻¹¹m=0.53A.

Ques. State the limitations of Bohr’s atomic model.

Ans. Here are the following limitations of Bohr’s atomic model:

(a) It does not indicate the structure and distribution of electrons in an atom.

(b) It could not account for the spiral nature of electrons.

Ques. Assume you are offered the possibility of repeating the alpha-particle diffusion experimentation using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is stable at temperatures below 14K.) What outcomes do you expect?

Ans.  In the alpha-particle diffusion experimentation, when a thin sheet of solid hydrogen is used in place of a gold foil, the scattering angle would not be large enough. The mass of hydrogen (1.67×10⁻²⁷kg) is less than that of the incident α -particles (6.64×10⁻²⁷ kilograms). As a result, the scattering particle has a mass larger than the target nucleus (hydrogen). So, if solid hydrogen is employed in the α-particle scattering experiment, the α-particles will not bounce back.

Ques. What fraction of tritium will remain after 25 years? Given the half-life of tritium as 12.5 years.

Ans. It is given that,

t = 25 years

T = 12.5 years 

N/N₀ = (1/2)^t/T

= (1/2)^25/12.5 

⇒N/N₀ = (1/2)²=1/4

⇒N/N₀ = 0.25, 

Which is the required fraction.

Ques. Calculate the kinetic energy and potential energy of an electron in the first orbit of a hydrogen atom. Given e = 1.6×10⁻¹9C and r = 0.53×10⁻¹⁰m.

Ans.  a) Kinetic Energy

K.E.= ke²/2r

⇒K.E.=(1.6×10⁻¹⁹)2×9×10⁹ / 2×0.53×10⁻¹⁰

⇒K.E.=21.74×10⁻¹⁹J

⇒K.E.=21.74×10⁻¹⁹ / 1.6×10⁻¹⁹ = 13.59eV

b) Potential Energy

P.E. = −ke² / r = −2K.E.

⇒ P.E. = −2×13.59=−27.18eV

Therefore, Kinetic energy is 13.59eV and Potential energy is −27.18eV.

Ques. Why is nuclear fusion not possible in the laboratory?

Ans. Nuclear fusion is not feasible in the laboratory as it is conducted at high temperatures. So you can not attain this in the laboratory.

Also check: P-n Junction

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Long Answer Question [3 Marks Questions]

Ques. The fission properties of ²³⁹₉₄ Pu are remarkably identical to those of 23592 U. The average energy emitted per fission is 180 MeV. How much energy in MeV is discharged if all the atoms in 1kg of pure ²³⁹₉₄ Pu experience fission?

Ans. Given that,

Average energy emitted per fission of ²³⁹₉₄ Pu, E_av=180 MeV 

Amount of pure ²³⁹₉₄ Pu, m = 1kg = 1000g

N_A= Avogadro number = 6.023×10²³ 

Mass number of ²³⁹₉₄ Pu = 239g

1 mole of ²³⁹₉₄Pu contains N_A atoms.

Therefore, 1kg of ²³⁹₉₄Pu contains {( N_A / Mass number)×m} atoms 

⇒ {(6.023×10²³) / 239} ×1000 = 2.52×10²⁴ atoms 

Thus, the total energy released during the fission of 1 kg of ²³⁹₉₄Pu: E = E_av×2.52×10²⁴ 

⇒E = 180×2.52×10²⁴

= 4.536 × 10²⁶ MeV

Therefore, 4.536 × 10²⁶ MeV is released if all the atoms in 1 kg of pure ²³⁹₉₄ Pu undergo fission.

Ques. Calculate the height of the potential barrier for a head-on collision of two deuterons. (Hint: The size of the possible wall is given by the Coulomb repulsion between the two deuterons when they touch each other. Assume that they can be taken as hard spheres of radius 2.0fm.)

Ans. If two deuterons collide head-on then, the distance between their centers is:

d = Radius of 1st deuteron + Radius of 2nd deuteron 

Radius of a deuteron nucleus = 2 f_m 

= 2 × 10⁻¹⁵ m

⇒ d = 2 × 10⁻¹⁵ + 2 × 10⁻¹⁵

= 4 × 10⁻¹⁵ m 

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10⁻¹⁹ C 

The potential energy of the two-deuteron system:

V = e² / 4πε₀d 

Where,

ε0 is the Permittivity of free space

1/4πε₀ = 9 × 10⁹Mm²c⁻²

⇒ V = [ 9 × 10 × (1.6 × 10⁻¹⁹)² / 4 × 10⁻¹⁵ ] J

⇒ V = [ 9 × 10 × (1.6 × 10⁻¹⁹)² / 4 × 10⁻¹⁵  × (1.6×10⁻¹⁹) ] eV

⇒ V = 360 keV 

Therefore, the height of the potential barrier of the two-deuteron system is 360 keV.

Ques. A difference of 2.3eV separates two energy levels in an atom. What is the radiation frequency when the atom transitions from the upper level to the lower level?

Ans. We are given the separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10⁻¹⁹

= 3.68 × 10⁻¹⁹ J

If v is the frequency of radiation emitted when the atom transits from the upper level to the lower level, the energy is given by,

E = hv

Where, h= Planck's constant = 6.62×10⁻⁴ Js

⇒ v = E / h

Substituting the given values, 

ν = 3.68 × 10⁻¹⁹ / 6.62×10⁻³²

= 6.62 × 10⁻³²

= 5.55 × 10¹⁴ Hz

Hence, the frequency of the radiation is found to be 5.55 × 10¹⁴ Hz.

Related Links:

Ques. The total energy of an electron in the first excited state of the hydrogen atom is about −3.4eV.

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) We are given,

Total energy of the electron, E = −3.4 eV

The kinetic energy of the electron is equal to the negative of the total energy.

⇒ K.E = −E

∴ K.E = −(−3.4) = + 3.4eV

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

b) We know that the potential energy (U) of the electron is found to be equal to the negative of twice its kinetic energy.

⇒ U = − 2 K.E

∴ U = − 2 × 3.4

= −6.8 eV

Hence, the potential energy of the electron in the given state is found to be −6.8 eV

c) We understand that the potential energy would rely on the connection point brought. Here, the potential energy of the connection point is considered zero. The system's potential energy value would also change by adjusting the reference point. Since we know that total energy is the sum of kinetic and potential energies, the system's total energy will also change.

Also check: Semiconductor Diode

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Very Long Answer Question [5 Marks Question]

Ques. A radioactive isotope has a half-life of X years. How long will it take the activity to reduce to 

a) 3.125

b) 1 of its original value

Ans. a) Let the half-life of the radioactive isotope be X years. The original amount of the radioactive isotope will be N₀. After decay, the amount of the radioactive isotope will be N. It is given that only 3.125 of N₀ remains after decay.

⇒ N / N₀ = 3.125 

It is known that, N / N₀ =e^−λt

Where λ is the Decay constant.

t is the Time.

⇒ e^−λt =1 / 32

⇒ −λt = ln 1− ln 32

⇒ −λt = 0 − 3.4567

⇒ t = 3.4567 / λ

It is known that, λ = 0.693 / X

⇒ t = 3.4567 / 0.693 / X = 5X

Therefore, the isotope will take about 5X years to reduce to 3.125 of its original value.

b) Suppose that after decay, the amount of the radioactive isotope is N.

It is given that only 1of N₀ remains after decay.

⇒ N / N₀ = 1 

It is known that N / N₀ = e^−λt

Where λ is the Decay constant. t is the Time.

⇒ e^−λt =1100 

⇒ −λt = ln 1 −ln 100

⇒ −λt = 0 − 4.6052

⇒ t = 4.6052λ

It is known that, λ = 0.693 / X

⇒ t = 4.6052 / 0.693 X = 6.645 X

Therefore, the isotope will take about 6.645 X years to reduce to 1 of its original value.

Ques. Answer the following question, which will help you better understand the difference between Thomson's and Rutherford's models.

1. 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?
2. 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?
3. 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?
4. In which model is it completely wrong to ignore multiple scattering for the calculation of the average angle of scattering of α-particles by a thin foil?

Ans. a) About the same;

The average angle of deflection of α-particles by a delicate gold foil indicated by Thomson's model is about the same size as Rutherford's model. And this is because the average angle was accepted in both these models.

b) Much less;

The possibility of diffusion of α-particles at angles greater than 90 degrees predicted by Thomson's model is much less than Rutherford's model.

c) We know that diffusion is large because of single crashes. The probabilities of a single wreck increase linearly with the number of target atoms. Since the number of target atoms increases with an increase in thickness, the collision probability would depend linearly on the thickness of the target.

d) Thomson's model;

It is wrong to ignore multiple scattering in Thomson's model to calculate the average angle of dispersion of α-particles by a thin foil. A single collision could cause very little deflection in this model. Hence, you can observe the average scattering angle and explain it only by considering multiple scattering.

Ques. The total energy of an electron in the first excited state of a hydrogen atom is −3.4eV. Calculate:

a) K.E. of the electron in this state.

b) P.E. of the electron in this state.

c) Which of the answers would change if zero of PE is changed? Justify your answer.

Ans.

1. We know that K.E = −E

∴ K.E = 3.4 eV

2. P.E = 2 × K.E

∴ P.E = 2 × 3.4 = 6.8 eV

100. If the zero of the P.E is changed, K.E will remain intact, but the P.E will alter, and so will the total energy.

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