1.

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 isA. 2 eVB. 1 eVC. 0.5 eVD. none of these

Answer» Correct Answer - 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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