Question

A free electron of $2.6eV$ 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 $=2.6+0=2.6eV$
Finally in first excited state of $H$ atom total energy $=-3.4eV$
Loss in total energy $=2.6-(-3.4)$
$=6eV$
It is emitted as photon
$\lambda =\frac{1240}{6}=206nm$
$f=\frac{3\times 10^8}{206\times 10^{-9}}=1.45\times 10^{15}Hz$
$=1.45\times 10^{9}Hz$