The recoil speed of a hydrogen atom after it emits a photon in going from $n = 5$ state to $n = 1$ state is

4.718 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

7.418 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4.178 m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

7.148 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $4.178 m s^{- 1}$
Momentum, $p = \frac{E}{c} = m v$
velocity, $v = \frac{δ E}{m c}$ , mass of hydrogen atom $m = 1.67 \times 10^{- 27} K g$

$E = \frac{- 13.6 Z^{2}}{n^{2}}$

$δ E = E_{5} - E_{1} = - 13.056 e V$

Putting the value in velocity expression we get $v = 4.175 m s^{- 1}$

Suggest Corrections

Similar questions

n = 2

n = 1

R_{\infty} = 1.1 \times 10^{7} m^{- 1}

h = 6.63 \times 10^{- 34} J s

n = 5

n = 1

n = 5 t o n = 1

R_{\infty} = 1.1 \times 10^{- 1}

h = 6.63 \times 10^{- 34} Js]

n = 5

n = 1

Join BYJU'S Learning Program