A hydrogen atom is in fifth excited state. When the electron jumps to ground state, the velocity of recoiling hydrogen atom is

Question

A hydrogen atom is in fifth excited state. When the electron jumps to ground state, the velocity of recoiling hydrogen atom is (A) 1.1 m s- 1 (B) 4.2 m s- 1 (C) 8.4ms- 1 (D) 11.2ms- 1

Tardigrade Admin · Accepted Answer

Δ E=13.6((1/12) - (1/62)) Δ E=13.22 eV p=(Δ E/c) ⇒ v=(Δ E/m C)=(13.22 × 1.6 × 1 0- 19/1.67 × 1 0- 27 × 3 × 1 08)=4.2 m s- 1

Q. A hydrogen atom is in fifth excited state. When the electron jumps to ground state, the velocity of recoiling hydrogen atom is

$1.1 m s^{- 1}$

$4.2 m s^{- 1}$

$8.4 m s^{- 1}$

$11.2 m s^{- 1}$

$Δ E = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{6 ^{2}})$
$Δ E = 13.22 e V$
$p = \frac{Δ E}{c}$
$\Rightarrow v = \frac{Δ E}{m C} = \frac{13.22 \times 1.6 \times 1 0 ^{- 19}}{1.67 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{8}} = 4.2 m s^{- 1}$

Q. A hydrogen atom is in fifth excited state. When the electron jumps to ground state, the velocity of recoiling hydrogen atom is

1.1ms−1

4.2ms−1

8.4ms−1

11.2ms−1

ΔE=13.6(121​−621​) ΔE=13.22eV p=cΔE​ ⇒v=mCΔE​=1.67×10−27×3×10813.22×1.6×10−19​=4.2ms−1

$1.1 m s^{- 1}$

$4.2 m s^{- 1}$

$8.4 m s^{- 1}$

$11.2 m s^{- 1}$

$Δ E = 13.6 (\frac{1}{1 ^{2}} - \frac{1}{6 ^{2}})$
$Δ E = 13.22 e V$
$p = \frac{Δ E}{c}$
$\Rightarrow v = \frac{Δ E}{m C} = \frac{13.22 \times 1.6 \times 1 0 ^{- 19}}{1.67 \times 1 0 ^{- 27} \times 3 \times 1 0 ^{8}} = 4.2 m s^{- 1}$