Question 188: Dual Nature of Matter and Radiation

Question

Surface of certain metal is first illuminated with light of wavelength $$\lambda $$
1
= 350 nm and then, by light of wavelength $$\lambda $$
2
= 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :
(Energy of photon n = $${{1240} \over {\lambda (in\,mm)}}$$eV)

Accepted Answer

Let speed of photon electron in first case is 2v
then in the second case speed is v.
For first case
$${{hc} \over {{\lambda _1}}} = \phi  + {1 \over 2}$$m(2v)
2
For second case,
$${{hc} \over {{\lambda _2}}} = \phi  + {1 \over 2}$$mv
2
$$ 	herefore $$   $${{{{hc} \over {{\lambda _1}}} - \phi } \over {{{hc} \over {{\lambda _2}}} - \phi }} = {{{1 \over 2}m 	imes 4{v^2}} \over {{1 \over 2}m{v^2}}}$$
$$ \Rightarrow $$   $${{{hc} \over {{\lambda _1}}} - \phi }$$ = 4$$\left( {{{hc} \over {{\lambda _2}}} - \phi } ight)$$
$$ \Rightarrow $$   $${{4hc} \over {{\lambda _2}}}$$ $$-$$ $${{hc} \over {{\lambda _1}}}$$ = 3$$\phi $$
$$ \Rightarrow $$   $$\phi $$ $$=$$ $${{hc} \over 3}\left( {{4 \over {{\lambda _2}}} - {1 \over {{\lambda _1}}}} ight)$$
$$ \Rightarrow $$   $$\phi $$ $$=$$  $${{1240} \over 3}\left( {{4 \over {540}} - {1 \over {350}}} ight)$$
$$ \Rightarrow $$   $$\phi $$ = 1.8 eV

Answer

1.8

Answer

2.5

Answer

5.6

Answer

1.4

Dual Nature of Matter and Radiation

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