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  4. Maximum velocity of photoelectrons is 3.5 × 106 m/s. If the specific charge of an electron is 1.75 × 1011 C/kg, stopping potential of the electron is

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Q. Maximum velocity of photoelectrons is $ 3.5 \times 10^6 \, m/s$. If the specific charge of an electron is $1.75 \times 10^{11} C/kg,$ stopping potential of the electron is

COMEDKCOMEDK 2005Dual Nature of Radiation and Matter

Solution:

As $eV_{0 } = K_{max} = \frac{1}{2} mv^{2}_{max} $
or, $V_{0 } = \frac{1}{2} \left(\frac{m}{e}\right) v_{max}^{2} $
Here,
$ v_{max} = 3.5 \times10^{6} \frac{m}{s} , \frac{e}{m} = 1.75 \times 10^{11} C/ kg $
$ \therefore \, V_{0} = \frac{\left(3.5 \times 10^{6}\right)^{2}}{2 \times 1.75 \times 10^{11}} = 35 V$