A hydrogen atom emits a photon corresponding to an electron transition from n=5 to n=1. The recoil speed of hydrogen atom is almost (mass of proton ≈ 1.6 × 10-27 kg )

Question

A hydrogen atom emits a photon corresponding to an electron transition from n=5 to n=1. The recoil speed of hydrogen atom is almost (mass of proton ≈ 1.6 × 10-27 kg ) (A)  10 ms-1  (B) 2 × 10-2 ms -1 (C) 4 m s-1 (D) 8 × 102 m s-1

Tardigrade Admin · Accepted Answer

The hydrogen atom before the transition was at rest. Therefore, from conservation of momentum

P_{H-atom} = P_{photon} = \frac{E _{radiated}}{c} = \frac{13.6 [ \frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}} ] e V}{c}

1.6 \times 1 0^{- 27} \times v

= \frac{13.6 [ \frac{1}{1 ^{2}} - \frac{1}{5 ^{2}} ] \times 1.6 \times 1 0 ^{- 19}}{3 \times 1 0 ^{6}}

\Rightarrow v = 4.352 m / s =\approx 4 m / s

Q. A hydrogen atom emits a photon corresponding to an electron transition from $n = 5$ to $n = 1$ . The recoil speed of hydrogen atom is almost
(mass of proton $\approx 1.6 \times 1 0^{- 27} k g$ )

$10 m s^{- 1}$

$2 \times 1 0^{- 2} m s^{- 1}$

$4 m s^{- 1}$

$8 \times 1 0^{2} m s^{- 1}$

Q. A hydrogen atom emits a photon corresponding to an electron transition from n=5 to n=1. The recoil speed of hydrogen atom is almost (mass of proton ≈1.6×10−27kg )

10ms−1

2×10−2ms−1

4ms−1

8×102ms−1

The hydrogen atom before the transition was at rest. Therefore, from conservation of momentum PH-atom​=Pphoton​=cEradiated ​​=c13.6[n12​1​−n22​1​]eV​ 1.6×10−27×v =3×10613.6[121​−521​]×1.6×10−19​ ⇒v=4.352m/s=≈4m/s

Q. A hydrogen atom emits a photon corresponding to an electron transition from $n = 5$ to $n = 1$ . The recoil speed of hydrogen atom is almost
(mass of proton $\approx 1.6 \times 1 0^{- 27} k g$ )

$10 m s^{- 1}$

$2 \times 1 0^{- 2} m s^{- 1}$

$4 m s^{- 1}$

$8 \times 1 0^{2} m s^{- 1}$