Atomic Physics

Q. When an electron jumps from a level n = 4 to n = 1, the momentum of the recoiled hydrogen atom will be

(a) 6.5 × 10^-27 kg ms^-1         

(b) 12.75 × 10^-19 kg ms^-1

(c) 13.6 × 10^-27 kg ms^-1      

(d)zero

Ans: (a)

Sol: Momentum of recoiled hydrogen atom , P = h/λ

$\large \frac{1}{\lambda} = R Z^2 [\frac{1}{n_1^2} – \frac{1}{n_2^2}]$

For hydrogen , Z = 1

$\large \frac{1}{\lambda} = R [\frac{1}{n_1^2} – \frac{1}{n_2^2}]$

As , P = h/λ

$\large P = h R [\frac{1}{n_1^2} – \frac{1}{n_2^2}]$

$\large P = 6.6 \times 10^{-34} \times 10^7 [\frac{1}{1^2} – \frac{1}{2^2}]$

= 6.5 × 10⁻²⁷ Kg m/s

Q. Let v₁ be the frequency of series limit of Lyman series, v₂ the frequency of the first line of Lyman series, and v₃ the frequency of series limit of Balmer series. Then which of the following is correct?

(a) v₁ – v₂ = v₃

(b) v₂ – v₁ = v₃

(c) v₃ = ½ (v₁ + v₂)

(d) v₂ + v₁ = v₃

Ans: (a)

Sol: Series Limit means Shortest possible wavelength .

A/c to question , ν = c[1/n₁²  −1/n₂²] , Where c = constant

ν₁ = c[1/1² − 1/∞] = c

For first line of Lyman Series ,

ν₂ = c[1/1² − 1/2² ] = 3c/4

ν₃ = c[1/2² − 1/∞] = c/4

Hence , ν₁ = ν₂ + ν₃
⇒ ν₁ − ν₂ =  ν₃

Q. Given: mass number of gold = 197, density of gold = 19.7g cm^-3, Avogadro’s number = 6 × 10²³. The radius of the gold atom is approximately:

(a) 1.5 × 10^-8 m

(b) 1.5 × 10^-9 m

(c) 1.5 × 10^-10 m

(d) 1.5 × 10^-12 m

Ans: (c)

Sol: Density , ρ = Mass/Volume

Volume V = mass/ρ = 197/19.7 = 10cm³
Volume of one atom = 10/6.023×10²³ = (5/3) × 10^-23 cm³

If R be the radius of atom then,

(4/3)πR³ = (5/3) ×10^-23

R³ = (5/4π)× 10^-23

R = 1.5×10^-10  m

Q. A stationary hydrogen atom of mass M emits a photon corresponding to the first line of Lyman series. If R is the Rydberg’s constant, the velocity that the atom acquired is

(a) 3Rh/4M

(b) Rh/4M

(c) Rh/2M

(d) Rh/M

Ans: (a)       

Sol: The first line of Lyman series ,

1/λ = R[1/1²  −1/2²] = R ×(3/4)

As , momentum P = h/λ =Mv

v = h/Mλ = (h/M)×R ×(3/4)

v = 3hR/4M

Q. The ratio of maximum to minimum possible radiation energy of Bohr’s hypothetical hydrogen atom is equal to

(a) 2

(b) 4

(c) 4/3

(d) 3/2

Ans: (c)

Sol: Energy of radiation b/w n₁ and n₂

E= 13.6[1/n₁²  −1/n₂²]

When n₁ = 1 & n₂ = 2 , E is minimum

E_min = 13.6[1−1/4] = 13.6×(3/4)

When n₁ = 1 & n₂ = ∞ , E is maximum

E_max = 13.6[1/1 −1/∞] = 13.6 eV

E_max/E_min  = 4/3

Q. Transitions between three energy levels in a particular atom give rise to three spectral lines of wavelengths, in increasing magnitudes, λ₁, λ₂ and λ₃. Which one of the following equations correctly relates λ₁, λ₂ and λ₃ ?

(a) λ₁ = λ₂– λ₃

(b) λ₁ = λ₃– λ₂

(c) 1/λ₁ = 1/ λ₂ + 1/ λ₃

(d)1/ λ₁ = 1/ λ₃– 1/λ₂

Ans: (c)

Sol: $\large \frac{1}{\lambda_1} = R Z^2 [\frac{1}{n_1^2} – \frac{1}{n_3^2} ]$ …(i)

$\large \frac{1}{\lambda_2} = R Z^2 [\frac{1}{n_1^2} – \frac{1}{n_2^2} ]$ …(ii)

$\large \frac{1}{\lambda_3} = R Z^2 [\frac{1}{n_2^2} – \frac{1}{n_3^2} ]$ …(iii)

From (i) , (ii) & (iii)

$\large \frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3} $

Q. If the first excitation potential of a hydrogen-like atom is V electron volt, then the ionization energy of this atom will be

(a) V electron volt

(b) 3V/4 electron volt

(c) 4V/3 electron volt

(d)cannot be calculated by given information

Ans: (c)

Sol: E −E/4 = V

3E/4 = V

E = 4V/3

Q. One of the lines in the emission spectrum of Li²⁺ has the same wavelength as that of the second line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is

(a) n = 4 → n = 2

(b) n = 8 → n = 2

(c) n = 8 → n = 4

(d) n = 12 → n = 6

Ans: (d)

Sol: $\large \frac{1}{\lambda} = R Z^2 [\frac{1}{n_1^2} – \frac{1}{n_2^2} ]$

For , Li , Z = 3

$\large \frac{1}{\lambda} = 9 R [\frac{1}{n_1^2} – \frac{1}{n_2^2} ]$

For hydrogen , 2nd line of Balmer series

$\large \frac{1}{\lambda} = R [\frac{1}{2^2} – \frac{1}{4^2} ]$

A/c to question

$\large = R [\frac{1}{2^2} – \frac{1}{4^2} ] = = 9 R [\frac{1}{n_1^2} – \frac{1}{n_2^2} ]$

$\large \frac{3}{36} = 9[\frac{1}{2^2} – \frac{1}{4^2} ] = = 9 R [\frac{1}{n_1^2} – \frac{1}{n_2^2} ]$

$ \large \frac{3}{36} = 9[\frac{1}{n_1^2} – \frac{1}{n_2^2}] $

$ \large \frac{1}{48} = [\frac{1}{n_1^2} – \frac{1}{n_2^2}] $

Satisfied for , n₁ = 6 , n₂  = 12

Q. In a hydrogen atom, the electron makes a transition from n = 2 to n = 1. The magnetic field produced by the circulating electron at the nucleus

(a) decreases 16 times

(b) increases 4 times

(c) decreases 4 times

(d) increases 32 times

Ans: (d)

Sol: Magnetic Field , $\large B = \frac{mu_0}{4\pi} \frac{e v}{r^2}$

$\large v = \frac{n h}{2\pi m r}$

As ,$\large r \propto n^2 $

$\large B \propto \frac{1}{n^5}$

Hence B increases by 32 times

Q: A particle of mass m moves in circular orbits with potential energy V(r) = F r , where F is positive constant and r is its distance from origin . Its energy are calculated using the Bohr model . If radius of the particle’s orbit is denoted by R and its speed and energy are denoted by v and E respectively , then for the n^th orbit (here h is the Planck’s constant)

(a) $R \propto n^{1/3}$ and $ v \propto n^{2/3}$

(b) $R \propto n^{2/3}$ and $ v \propto n^{1/3}$

(c) $ E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

(d) $ E = 2 (\frac{n^2 h^2 F^2}{4\pi^2 m})^{1/3}$

Ans: (b , c)

Solution: Force $|F| = |-\frac{dV}{dr}|$

$F = \frac{m v^2}{r} $ …(i)

$m v r = \frac{n h}{2 \pi}$ ….(ii)

From (i) & (ii)

$F = \frac{m}{r} (\frac{n h}{2 \pi m r})^2$

$F = \frac{m n^2 h^2}{4 \pi^2 m^2 r^3}$

$F = \frac{ n^2 h^2}{4 \pi^2 m r^3}$

Since F = constant

$r^3 \propto n^2 $

$r \propto n^{2/3} $

$v = \frac{n h}{2 \pi m r} $

$ v \propto \frac{n}{n^{2/3}} $

$ v \propto n^{1/3} $

$ K.E = \frac{1}{2}m v^2 = \frac{F r}{2} $

Total Energy = P.E + K.E

$T.E = F r + \frac{F r}{2}$

$T.E = \frac{3 F r}{2}$

$T.E = E = \frac{3F}{2} \times (\frac{n^2 h^2}{4 \pi^2 r n F})^{1/3} $

$E = \frac{3}{2} (\frac{n^2 h^2 F^2}{4 \pi^2 m})^{1/3}$