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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
Infrared region
B
X-ray region
C
Ultraviolet region
D
Visible region
Detailed Solution
Einstein's photoelectric equation: $E_{incident} = W + K_{max}$, with $K_{max} = eV_0$ where $V_0$ is the stopping potential.
$\dfrac{hc}{\lambda} = 6.2\,e + 5\,e = 11.2$ eV
$\lambda = \dfrac{hc}{11.2 \times 1.6 \times 10^{-19}}$
$\lambda = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{11.2 \times 1.6 \times 10^{-19}}$
$\lambda = 1.1 \times 10^{-7}$ m $= 110$ nm
Visible light lies between about 400 nm and 700 nm, and the ultraviolet region extends from about 400 nm down to about 10 nm.
Hence this wavelength lies in the ultraviolet region.
$\dfrac{hc}{\lambda} = 6.2\,e + 5\,e = 11.2$ eV
$\lambda = \dfrac{hc}{11.2 \times 1.6 \times 10^{-19}}$
$\lambda = \dfrac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{11.2 \times 1.6 \times 10^{-19}}$
$\lambda = 1.1 \times 10^{-7}$ m $= 110$ nm
Visible light lies between about 400 nm and 700 nm, and the ultraviolet region extends from about 400 nm down to about 10 nm.
Hence this wavelength lies in the ultraviolet region.