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

Question

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 (A) 7 volt (B) 3.5 volt (C) 10.5 volt (D) 35 volt

Tardigrade Admin · Accepted Answer

As eV0 = Kmax = (1/2) mv2max or, V0 = (1/2) ((m/e)) vmax2 Here, vmax = 3.5 ×106 (m/s) , (e/m) = 1.75 × 1011 C/ kg ∴ V0 = ((3.5 × 106)2/2 × 1.75 × 1011) = 35 V

Q. Maximum velocity of photoelectrons is $3.5 \times 1 0^{6} m / s$ . If the specific charge of an electron is $1.75 \times 1 0^{11} C / k g,$ stopping potential of the electron is

7 volt

3.5 volt

10.5 volt

35 volt

As $e V_{0} = K_{ma x} = \frac{1}{2} m v_{ma x}^{2}$
or, $V_{0} = \frac{1}{2} (\frac{m}{e}) v_{ma x}^{2}$
Here,
$v_{ma x} = 3.5 \times 1 0^{6} \frac{m}{s}, \frac{e}{m} = 1.75 \times 1 0^{11} C / k g$
$∴ V_{0} = \frac{( 3.5 \times 1 0 ^{6} ) ^{2}}{2 \times 1.75 \times 1 0 ^{11}} = 35 V$

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