If maximum velocity with which an electron can be emitted from a photo cell is 4 × 108 cm/s, the stopping potential is (mass of electron = 9 × 10–31 kg)

To find the stopping potential, we can use the equation:

eV = 1/2 mv^2

where e is the charge of an electron (1.6 × 10^-19 C), V is the stopping potential, m is the mass of the electron (9 × 10^-31 kg), and v is the maximum velocity of the electron (4 × 10^8 cm/s).

First, let's convert the maximum velocity from cm/s to m/s:

v = 4 × 10^8 cm/s = 4 × 10^6 m/s

Now we can substitute the values into the equation:

(1.6 × 10^-19 C) V = 1/2 (9 × 10^-31 kg) (4 × 10^6 m/s)^2

Simplifying the equation:

V = (1/2) (9 × 10^-31 kg) (4 × 10^6 m/s)^2 / (1.6 × 10^-19 C)

V = (1/2) (9 × 10^-31 kg) (16 × 10^12 m^2/s^2) / (1.6 × 10^-19 C)

V = (1/2) (9 × 16) × (10^-31 kg) (10^12 m^2/s^2) / (1.6 × 10^-19 C)

V = (1/2) (144) × (10^-19 kg m^2/s^2) / (1.6 × 10^-19 C)

V = 72 × (10^-19 kg m^2/s^2) / (1.6 × 10^-19 C)

V = 45 × 10^-19 kg m^2/s^2 / C

V = 45 × 10^-19 V

Therefore, the stopping potential is 45 × 10^-19 volts.