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Question: A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is inci...

A light of wavelength 600 nm is incident on a metal surface. When light of wavelength 400 nm is incident , the maximum kinetic energy of the emitted photoelectrons is double. The work function of the metal is :

A

1.03eV

B

2.11eV

C

4.14eV

D

2.43 eV

Answer

1.03eV

Explanation

Solution

: Here λ=600 nm,λ′=400 nm,kmax ′=2Kmax⁡\lambda = 600 \mathrm {~nm} , \lambda ^ { \prime } = 400 \mathrm {~nm} , k _ { \text {max } } ^ { \prime } = 2 K _ { \max }

According to Einstein’s photoelectric equation

Kmax⁡=hcλ−ϕ0…….(i)K _ { \max } = \frac { h c } { \lambda } - \phi _ { 0 } \ldots \ldots . ( i ) and 2Kmax⁡=hcλ′−ϕ0……(ii)2 K _ { \max } = \frac { h c } { \lambda ^ { \prime } } - \phi _ { 0 } \ldots \ldots ( i i )

Dividing (ii) by (i) we get

2=(hcλ′)−ϕ0(hcλ)−ϕ02 = \frac { \left( \frac { h c } { \lambda ^ { \prime } } \right) - \phi _ { 0 } } { \left( \frac { h c } { \lambda } \right) - \phi _ { 0 } }

Or 2hcλ−2ϕ0=hcλ′−ϕ0\frac { 2 h c } { \lambda } - 2 \phi _ { 0 } = \frac { h c } { \lambda ^ { \prime } } - \phi _ { 0 }

Or hc(2λ−1λ′)=ϕ0(h c \left( \frac { 2 } { \lambda } - \frac { 1 } { \lambda ^ { \prime } } \right) = \phi _ { 0 } \quad ( Take hc=1240eVn m)h c = 1240 \mathrm { eV } n \mathrm {~m} )

∴ϕ0=1240eVnm(2600 nm−1400 nm)=1.03ev\therefore \phi _ { 0 } = 1240 \mathrm { eV } \mathrm { nm } \left( \frac { 2 } { 600 \mathrm {~nm} } - \frac { 1 } { 400 \mathrm {~nm} } \right) = 1.03 \mathrm { ev }