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The photoelectric threshold for some mat...

The photoelectric threshold for some material is 200 nm. The material is irradiated with radiations of wavelength 40 nm. The maximum kinetic energy of the emitted photoelectrons is

A

2 eV

B

1 eV

C

0.5 eV

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
D
Work function `W_0=(hc)/(lamda_0)`
`implies=W_0=(2xx10^(-25))/(2xx10^(-7))=10^(-18)J=(10^(-18))/(1.6xx10^(-19))eV=6.25eV`
energy of incident ratiation is `E=(hc)/(lamda)=31.25eV`
KE of photoelectrons `=E-W_0=25eV`
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