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Physics Question on Photoelectric Effect

The photoelectric threshold for a certain metal is 3600 A˚3600\, \mathring A. The maximum energy of the ejected photoelectrons by radiation of 2000 A˚2000\, \mathring A is (Given h=6.62×10−34J sh = 6.62 \times 10^{-34} J \, s )

A

1.86 eV

B

2.76 eV

C

7.26 eV

D

5.76 eV

Answer

2.76 eV

Explanation

Solution

Here,
Threshold wavelength, λ0=3600\lambda_0 = 3600
\hspace55mm = 3600 \times 10^{-10} m
\hspace55mm = 36 \times 10^{-8} m
Incident wavelength, λ=2000=2000×10−10m\lambda = 2000 = 2000 \times 10^{-10} m
\hspace55mm = 20 \times 10^{-8} m
According to Einstein's photoelectric equation
Kmax=hcλ−hcλ0=hc[1λ−1λ0]K_{max} = \frac{hc}{\lambda} - \frac{hc}{\lambda _{0}} = hc \left[\frac{1}{\lambda } - \frac{1}{\lambda_{0} }\right]
=6.62×10−34×3×108[120×10−8−136×10−8]= 6.62 \times 10^{-34} \times 3 \times 10^{8} \left[\frac{1}{20\times 10^{-8}} - \frac{1}{36 \times 10^{-8}}\right]
=6.62×10−34×3×10810−8[120−136]= \frac{6.62 \times 10^{-34} \times 3\times 10^{8}}{10^{-8}} \left[\frac{1}{20} - \frac{1}{36}\right]
=6.62×10−34×3×10810−8[9−5180]= \frac{6.62 \times 10^{-34}\times 3 \times 10^{8}}{10^{-8}} \left[ \frac{9-5}{180}\right]
=6.62×10−34×3×108×4180×10−8= \frac{6.62 \times 10^{-34} \times 3\times 10^{8}\times 4}{180\times 10^{-8}}
=6.62×10−34×3×108×4180×10−8×1.6×10−19eV=2.76eV= \frac{6.62 \times 10^{-34}\times 3\times 10^{8}\times 4}{180 \times 10^{-8} \times 1.6 \times 10^{-19}} eV = 2.76 eV