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The work function of a surface of a pho...

The work function of a surface of a photosensitive material is `6.2 eV`. The wavelength of the incident radiation for which the stopping potential is `5 V` lies in the

A

visible region

B

infrared region

C

X- ray region

D

ultraviolet region

Text Solution

Verified by Experts

The correct Answer is:
D
`hv_0` = work function `phi_0=6.2eV`
Stopping potential , `V_0=5`
`K.E._(max)=eV_0=5eV`
From Einstein.s photoelectric equation
`hv=hv_0+K.E._(max)`
`hv=6.2eV+5.0eV`
`(hc)/lamda=11.2 eV`
`lamda=(4.136xx10^(-15)xx3xx10^8(eV)m)/(11.2eV)`
`lamda=1.11xx10^(-7)m`
This wavelength lies in region of ultraviolet rays.
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Knowledge Check

  • For a photocell, the work function is phi and the stopping potential is V_(s) . The wavelength of the incident radiation is

    A
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