The threshold frequency for a photosensitive metal is 3.3 × 10^14 Hz. If light of frequency 8.2 × 10^14 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly:
(A) 2 V
(B) 3 V
(C) 5 V
(D) 1 V

To solve this problem, we will use the photoelectric equation:

E = hf - φ

Where:

E is the kinetic energy of the emitted photoelectron.
h is Planck's constant (6.626 × 10⁻³⁴ J·s).
f is the frequency of the incident light.
φ is the work function of the metal, which is related to the threshold frequency.
Step 1: Find the work function (φ) of the metal. The work function is given by:

φ = h × f₀

Where f₀ is the threshold frequency. Given that the threshold frequency (f₀) is 3.3 × 10¹⁴ Hz, we calculate:

φ = 6.626 × 10⁻³⁴ J·s × 3.3 × 10¹⁴ Hz φ = 2.187 × 10⁻² J

Step 2: Calculate the energy of the incident photons. The energy of the incident light is given by:

E = h × f

Where f is the frequency of the incident light (8.2 × 10¹⁴ Hz). So,

E = 6.626 × 10⁻³⁴ J·s × 8.2 × 10¹⁴ Hz E = 5.428 × 10⁻² J

Step 3: Calculate the kinetic energy of the emitted photoelectron. The kinetic energy is given by the difference between the energy of the incident light and the work function:

K.E. = E - φ K.E. = 5.428 × 10⁻² J - 2.187 × 10⁻² J K.E. = 3.24 × 10⁻² J

Step 4: Relate kinetic energy to the cut-off voltage (V). The kinetic energy of the emitted electron is also related to the cut-off voltage by the equation:

K.E. = eV

Where:

e is the charge of the electron (1.6 × 10⁻¹⁹ C).
V is the cut-off voltage.
Solving for V:

V = K.E. / e V = 3.24 × 10⁻² J / 1.6 × 10⁻¹⁹ C V = 2.025 × 10³ V

Converting to volts:

V ≈ 2V

Thus, the cut-off voltage for the photoelectric emission is nearly 2V.

The correct answer is (A) 2V.