Question

A free electron of $2.6\text{eV}$ 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. $(h=6.6\times {10}^{-34}\text{Js})$

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

Answer 1

The correct option is D $1.45\times {10}^9\text{MHz}$

Initially total energy of free electron is given by $2.6\text{eV}$

Free electron collides with $H^{+}$ ion produces hydrogen atom in first excited state and emits photon.

Finally in first excited state of $H$ atom total energy is $-3.4\text{eV}$

Loss of energy during this process is $E_{\text{photon}}=2.6-(-3.4)=6\text{eV}$

Frequency of photon emitted is ${\nu }_{\text{photon}}=\frac{E_{\text{photon}}}{h}=\frac{6\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}=1.45\times {10}^{15}\text{Hz}=1.45\times {10}^9\text{MHz}$