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

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

## Let v1 be the frequency of series limit of Lyman series, v2 the frequency of the first line of Lyman series, and v3 the….

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

## Given: mass number of gold = 197, density of gold = 19.7g cm-3, Avogadro’s number = 6 × 10^23. The radius of the gold atom is approximately

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

## A stationary hydrogen atom of mass M emits a photon corresponding to the first line of Lyman series. If R is the…..

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

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

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

Emin = 13.6[1−1/4] = 13.6×(3/4)

When n1 = 1 & n2 = ∞ , E is maximum

Emax = 13.6[1/1 −1/∞] = 13.6 eV

Emax/Emin  = 4/3

## Transitions between three energy levels in a particular atom give rise to three spectral lines of wavelengths…..

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)

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

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

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

## One of the lines in the emission spectrum of Li2+ has the same wavelength as that of the second line of Balmer series….

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 , n = 12

## 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

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

## A particle of mass m moves in circular orbits with potential energy V(r) = F r , where F is positive constant ….

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

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

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

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

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