The threshold frequency for a photosensitive metal is 3.3×104Hz. If light of frequency 8.2×1014Hz is incident on this metal. The cut-off voltage for the photoelectric emission is nearly

To solve this problem, we need to use the photoelectric effect equation, which is given by:

E = hf - Φ

where: E is the energy of the emitted electron, h is Planck's constant (6.63 x 10^-34 Js), f is the frequency of the incident light, and Φ is the work function of the metal.

The work function can be calculated using the threshold frequency (f0), which is the minimum frequency of light required to emit electrons. It is given by:

Φ = hf0

Substituting the given threshold frequency (3.3 x 10^4 Hz) into this equation gives:

Φ = (6.63 x 10^-34 Js) * (3.3 x 10^4 Hz) = 2.18 x 10^-29 J

Next, we can calculate the energy of the emitted electron using the frequency of the incident light (8.2 x 10^14 Hz):

E = hf = (6.63 x 10^-34 Js) * (8.2 x 10^14 Hz) = 5.43 x 10^-19 J

The energy of the emitted electron is also equal to the product of the electron's charge (e = 1.6 x 10^-19 C) and the cut-off voltage (V). Therefore, we can solve for V:

E = eV V = E / e = (5.43 x 10^-19 J) / (1.6 x 10^-19 C) = 3.39 V

Therefore, the cut-off voltage for the photoelectric emission is nearly 3.39 V.