Question

A beam of electron accelerated with $4.64V$ is passed through a tube containing 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. $3.75\times {10}^{4}c{m}^{-1}$
• B. $2.1\times {10}^{7}c{m}^{-1}$
• C. $3.75\times {10}^6{m}^{-1}$
• D. None of these

Answer 1

Answer: (A), (C)

The correct options are
A $3.75\times {10}^{4}c{m}^{-1}$
C $3.75\times {10}^6{m}^{-1}$

Energy of the accelerated electron = 4.64 eV = $1.602\times {10}^{-19}\times 4.64$ J $=7.43\times {10}^{-19}J$

Formula: $\Delta E=h\upsilon =\frac{hc}{\lambda }$

where,

$h$ is the Planck's constant

$\upsilon$ is the frequency of the emitted light

$c$ is the speed of the emitted light

$\lambda$ is the wavelength of the emitted light

Wave number (or the number of waves in a unit distance) = $\frac{1}{\lambda }$ = $\frac{\Delta E}{hc}$ = $\frac{7.43\times {10}^{-19}}{(6.625\times {10}^{-34})\times (3\times {10}^8)}$

= $3.74\times {10}^6{m}^{-1}$

= $3.74\times {10}^4{cm}^{-1}$

Hence, the answer is A and C.