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Find the difference of kinetic energies ...

Find the difference of kinetic energies of photoelectrons emitted from a surface by light of wavelength 2500Å and 5000Å. `h=6.62xx10^(-34)Js`.

A

1.61 eV

B

2.47 eV

C

3.96 eV

D

`3.96xx10^(-19)` eV

Text Solution

Verified by Experts

The correct Answer is:
B
As, `DeltaE=(hc)/(lamda_(1))-(hc)/(lamda_(2))=(hc(lamda_(2)-lamda_(1)))/(lamda_(1)lamda_(2))` (in eV)
`=(6.62xx10^(-34)xx3xx10^(8)xx(5000-2500)xx10^(-10))/(2500xx5000xx10^(-20)xx1.6xx10^(-19))`
`=2.47eV`
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