Peak emission from a black body at a certain temperature occures at a wavelength lambda. On increasing its temperature , total radiation emitted is increasing 16 times . At the initial temperature when peak radiation from the black body is incident on a metal surface , it does not cause photoemission from surface . After the increasing the peak radiation from black body caused photoemission. to bring these photoelectrons to rest, a potential equivalent to the excitation energy between `n=2` to `n=3` Bohr levels of hydrogen atom is required. if eork function of metal is `2.24eV`, then value of lambda is [hc=12400eV-Å]

Options

A. `3000Å`
B. `6000Å`
C. `9000Å`
D. `12000Å`

Answer

B

Explanation

`E=eAsigmaT^(4)RightarrowEpropT^(4)RightarrowE_(1)/(16E_(1))=(T_(1)/T_(2))^(4)RightarrowT_(2)=2T_(1)`
From wein's displacement law `lambda_(m)R=b`
` Rightarrow lambda_(2)T_(1)=lambda_(2)T_(2)Rightarrowlambda_(2)=lambda_(1)/2=lambda/2`
` KE_(max)=eV_(s)=13.6(1/4-1/9)=1.8eV`
` i.e(hc)/lambda_(2)=eV_(s)+W=1.89+2.24=4.13eV`
` Rightarrowlambda_(2)=(12400)/(4.130)=3000A^(0) Rightarrow lambda=6000A^(0)`