Question 204: Dual Nature of Matter and Radiation

Question

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to:  ($${{{hc} \over e}}$$ = 1240 nm eV)

Accepted Answer

$${{hc} \over {{\lambda _1}}} = \phi  + e$$V
1
. . . (i)
$${{hc} \over {{\lambda _2}}} = \phi  + e$$V
2
. . . (ii)
(i) $$-$$ (ii)
hc$$\left( {{1 \over {{\lambda _1}}} - {1 \over {{\lambda _2}}}} ight)$$ = e(V
1
$$-$$ V
2
)
$$ \Rightarrow $$  V
1
$$-$$ V
2
= $${{hc} \over e}$$$$\left( {{{{\lambda _2} - {\lambda _1}} \over {{\lambda _1} - {\lambda _2}}}} ight)$$
= (1240nm $$-$$ V) $${{100nm} \over {300nm 	imes 400nm}}$$
= 1V

Answer

0.5 V

Answer

1.0 V

Answer

2.0 V

Answer

1.5 V

Dual Nature of Matter and Radiation

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to: (${{{hc} \over e}}$ = 1240 nm eV)

Choose the correct answer: