JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 10)
A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive 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?
0.96 V
1.25 V
0.24 V
1.5 V
Explanation
$$k{E_{\max }} = {{hc} \over {{\lambda _i}}} + \phi $$
or $$e{V_o} = {{hc} \over {{\lambda _i}}} + \phi $$
when $$\lambda$$i = 670.5 nm ; Vo = 0.48
when $$\lambda$$i = 474.6 nm ; Vo = ?
So,
$$e(0.48) = {{1240} \over {670.5}} + \phi $$ ..... (1)
$$e({V_o}) = {{1240} \over {474.6}} + \phi $$ .....(2)
(2) $$-$$ (1)
$$e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV$$
$${V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)$$ Volts
Vo = 0.48 + 0.76
Vo = 1.24 V $$ \simeq $$ 1.25 V
or $$e{V_o} = {{hc} \over {{\lambda _i}}} + \phi $$
when $$\lambda$$i = 670.5 nm ; Vo = 0.48
when $$\lambda$$i = 474.6 nm ; Vo = ?
So,
$$e(0.48) = {{1240} \over {670.5}} + \phi $$ ..... (1)
$$e({V_o}) = {{1240} \over {474.6}} + \phi $$ .....(2)
(2) $$-$$ (1)
$$e({V_o} - 0.48) = 1240\left( {{1 \over {474.6}} - {1 \over {670.5}}} \right)eV$$
$${V_o} = 0.48 + 1240\left( {{{670.5 - 474.6} \over {474.6 \times 670.5}}} \right)$$ Volts
Vo = 0.48 + 0.76
Vo = 1.24 V $$ \simeq $$ 1.25 V
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