Question

A beam of electrons accelerated with $4.64V$ was passed through a tube having mercury vapours. As a result of absorption, electronic changes occurred with mercury atoms and light was emitted. If the full energy of single electron was converted into light, what was the wave number of emitted light?

• A. $4.75\times {10}^5{cm}^{-1}$.
• B. $3.75\times {10}^4{cm}^{-1}$.
• C. $4.75\times {10}^4{cm}^{-1}$.
• D. $3.75\times {10}^6{cm}^{-1}$.

Answer 1

Answer: (B)

$3.75\times {10}^4{cm}^{-1}$.

Energy absorbed$=4.64eV=4.64\times 1.6\times {10}^{-19}J=7.424\times {10}^{-n}J$

$\because \frac{hc}{\lambda }=energy$

$\therefore$ wave number $=\frac{1}{\lambda }=\frac{Energyobserved}{hc}=\frac{7.424\times {10}^{-19}}{6.626\times {10}^{-34}\times 3\times {10}^8}{m}^{-1}=3.75\times {10}^4$ $c{m}^{-1}$