A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photosensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?
(1) 0.96 V
(2) 1.25 V
(3) 0.24 V
(4) 1.5 V
Correct answer: (2) 1.25 V
Explanation:
kEmax = \(\frac{hc}{\lambda_i} + \phi\)
or eVo = \(\frac{hc}{\lambda_i} + \phi\)
when λi = 670.5 nm ; Vo = 0.48
when λi = 474.6 nm ; Vo = ?
So, e(0.48) = \(\frac{1240}{670.5} + \phi\) ......(i)
eVo = \(\frac{1240}{474.6} + \phi\) ......(ii)
(ii) - (i)
e(Vo - 0.48) = 1240 \(\Big(\frac{1}{474.6} - \frac{1}{670.5}\Big)\)eV
Vo = 0.48 + 1240 \(\Big(\frac{670.5 - 474.6}{474.6 \times 670.5}\Big)\) Volts
Vo = 0.48 + 0.76
Vo = 1.24 V ≅ 1.25 V