Light of wavelength 0.6pm from a sodium lamp falls on a photo cell and causes the emission o{ photo-electrons for which the stopping potential is 0.5 V With light of wavelength 0.40 pm from a mercu4r vapour lamp the stopping potential is 1'5 V then work-function in electron volt of the photo cell surface is

A
0.75 eV
B
1.5 eV
C
3 eV
D
3.5 eV

Solution:

we know $\frac{hc}{λ} - W_{0} = {eV}_{s}$
$\frac{hc}{0 \cdot 6 \times 10^{- 6}} = e (0 \cdot 5) + W_{0} and \frac{hc}{0 \cdot 4 \times 10^{- 6}} = e (1 \cdot 5) + W_{0} Dividing eq . (1) and (2), we ge \frac{0 \cdot 4}{0 \cdot 6} = \frac{0 \cdot 5 e + W_{0}}{1 \cdot 5 e + W_{0}} or W_{0} = 1 \cdot 5 e joule or \frac{W_{0}}{e} = 1 \cdot 5 eV$

Solution:

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