1.

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 theA. Infrared regionB. X-ray regionC. Ultraviolet regiD. Visible region

Answer» Correct Answer - C
`E_("incident")=W+K_("max"),K_("max")=eV_(0)`
`hv=hv_(0)+eV_(0)," stopping potential "=V_(0)`
`(hc)/(lambda)=6.2e+5e`
`lambda=(hxxc)/(11.2xx1.6xx10^(-19))=(6.63xx10^(-34)xx3xx10^(8))/(11.2xx1.6xx10^(-19))`
`=1.1xx10^(-7)m`


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