Light of wavelength $4000\ \mathring{\mathrm{A}}$ falls on a photosensitive metal and a negative $2V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 imes 10^{- 34}Js,\ e = 1.6 imes 10^{- 19}C,\ c = 3 imes 10^{8}ms^{- 1})$

Question

Light of wavelength $4000\ \mathring{\mathrm{A}}$ falls on a photosensitive metal and a negative $2V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 	imes 10^{- 34}Js,\ e = 1.6 	imes 10^{- 19}C,\ c = 3 	imes 10^{8}ms^{- 1})$

Dr. Abubacker · Accepted Answer

(a) Energy of incident light $E(eV) = \frac{12375}{4000} = 3.09\ eV$
Stopping potential is $- 2V\ $so $K_{\max} = 2\ eV$
Hence by using $E = W_{0} + K_{\max};W_{0} = 1.09\ eV = 1.1\ eV$

Q. Light of wavelength $4000\ \mathring{\mathrm{A}}$ falls on a photosensitive metal and a negative $2V$ potential stops the emitted electrons. The work function of the material (in $eV$) is approximately $(h = 6.6 \times 10^{- 34}Js,\ e = 1.6 \times 10^{- 19}C,\ c = 3 \times 10^{8}ms^{- 1})$

Explanation

Reviewed by Dr. Abubacker

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