11. ⇒ (MHT CET 2023 9th May Evening Shift )

The force acting on the electron in hydrogen atom (Bohr' theory) is related to the principle quantum number ' $n$ ' as

A. $n^{4}$

B. $n^{- 4}$

C. $n^{2}$

D. $n^{- 2}$

Correct Option is (B)

The centripetal force of the rotating electron is given by $F = \frac{m v^{2}}{r}$

But, according to Bohr,

$v \propto \frac{1}{n}$ and $r \propto n^{2}$

i.e. $F \propto \frac{v^{2}}{r^{2}}$

$\Rightarrow F \propto \frac{1}{n^{2} n^{2}}$

$F \propto \frac{1}{n^{4}}$

12. ⇒ (MHT CET 2023 9th May Evening Shift )

The wavelength of light for the least energetic photons emitted in the Lyman series of the hydrogen spectrum is nearly [Take $hc = 1240 eV$ - $nm$ , change in energy of the levels $= 10.2 eV$ ]

A. $150 nm$

B. $122 nm$

C. $102 nm$

D. $82 nm$

Correct Option is (B)

For least energetic photon emitted in Lyman series, $E = E_{2} - E_{1} = 10.2 eV$

$\begin{aligned} λ = \frac{hc}{E} & = \frac{1240}{10.2} nm \\ = 121.57 nm \approx 122 nm \end{aligned}$

13. ⇒ (MHT CET 2023 9th May Morning Shift )

For an electron moving in the $n^{th}$ Bohr orbit the deBroglie wavelength of an electron is

A. $n π r$

B. $\frac{π r}{n}$

C. $\frac{nr}{2 π}$

D. $\frac{2 π r}{n}$

Correct Option is (D)

From de Broglie's hypothesis,

$λ = \frac{h}{p_{n}} = \frac{h}{mv} .... (i)$

and from Bohr's atomic model,

$L = \frac{nh}{2 π}$

Also,

$L = {mvr}_{n}$

Due to quantization of angular momentum, we can write,

$\begin{aligned} \frac{nh}{2 π} = {mvr}_{n} \\ ∴ v = \frac{nh}{2 π mr} .... (ii) \end{aligned}$

putting (ii) into (i), we get,

$∴ λ = \frac{h}{m (nh / 2 π mr)} = \frac{2 π r}{n}$

14. ⇒ (MHT CET 2023 9th May Morning Shift )

If an electron in a hydrogen atom jumps from an orbit of level $n = 3$ to orbit of level $n = 2$ , then the emitted radiation frequency is (where R = Rydberg's constant, C = Velocity of light)

A. $\frac{3 RC}{27}$

B. $\frac{RC}{25}$

C. $\frac{8 RC}{9}$

D. $\frac{5 RC}{36}$

Correct Option is (D)

$\begin{array}{ll} \frac{1}{λ} = R [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}] \\ ∴ & \frac{1}{λ} = [\frac{1}{2^{2}} - \frac{1}{3^{2}}] = \frac{5}{36} R \\ ∴ & f = \frac{c}{λ} = \frac{5}{36} Rc \end{array}$

15. ⇒ (MHT CET 2021 21th September Evening Shift )

Ratio centripetal acceleration for an electron revolving in 3^rd and 5^th Bohr orbit of hydrogen atom is

A. 425 : 18

B. 625 : 81

C. 125 : 27

D. 221 : 36

Correct Option is (B)

Centripetal acceleration $a = \frac{v^{2}}{r}$

For hydrogen atom, $V \propto \frac{1}{n}$ and $r \propto n^{2}$

$∴ a \propto \frac{1}{n^{4}} ∴ \frac{a_{2}}{a_{5}} = {(\frac{5}{3})}^{4} = \frac{625}{81}$

⇒Structure of Atom and Nuclei