Question

A free electron of $2.6 \,eV$ energy collides with a $H ^{+}$ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. $\left( h =6.6 \times 10^{-34} Js \right)$

• A. $1.45 \times 10^{16} MHz$
• B. $0.19 \times 10^{15} MHz$
• C. $1,45 \times 10^{9} MHz$
• D. $9.0 \times 10^{27} MHz$

Answer 1

Answer: (C)

For every large distance P.E. $=0$
$\&$ total energy $=2.6+0=2.6\, eV$
Finally in first excited state of $H$ atom total energy $=-3.4 \,eV$
Loss in total energy $=2.6-(-3.4) $
$=6 \,eV $
It is emitted as photon
$\lambda =\frac{1240}{6}=206\, nm $
$f =\frac{3 \times 10^{8}}{206 \times 10^{-9}}=1.45 \times 10^{15} Hz$
$=1.45 \times 10^{9} Hz$