Question

A stationary $He^+$ ion emitted a photon corresponding to the first line of the Lyman series. That photon liberated a photo electron from a stationary $H$ atom in ground state. What is the velocity of photo electron ?
$RH = 109678 \,cm^{-1}$

• A. $9.1 × 10^8 \, cm\, sec^{-1}$
• B. $5.1 × 10^8 \, cm\, sec^{-1}$
• C. $6.1 × 10^8 \, cm\, sec^{-1}$
• D. $3.1 × 10^8 \, cm\, sec^{-1}$

Answer 1

Answer: (D)

Energy of photon liberated from $He ^{+}$during emission of first line of Lyman series
$= h C \cdot R _{ H } Z ^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$
$=6.625 \times 10^{-27} \times 3 \times 10^{10} \times 109678 \times 2^{2}\left[\frac{3}{4}\right] $
$=6.54 \times 10^{-11}$ erg
This energy is used in liberating electron from $H$ atom from ground state, therefore,
$6.54 \times 10^{-11}= E _{1}$ of $H +\frac{1}{2} mu ^{2}$
$=13.6 \times 1.602 \times 10^{-12}+\frac{1}{2} mu ^{2}$
$\frac{1}{2} mu ^{2}=6.654 \times 10^{-11}-2.179 \times 10^{-11}$
$=4.361 \times 10^{-11} erg$
$u ^{2}=\frac{4.361 \times 10^{-11} \times 2}{9.108 \times 10^{-28}}$
$u =3.09 \times 10^{8} cm\, sec ^{-1}$