Question

The photoelectric threshold wavelength of silver is 3250 × 10^–10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536 × 10^–10 m is
(Given h = 4.14 × 10^–15 eVs and c = 3 × 10⁸ ms^–1)

• A. ≃ 6 × 10⁵ ms^–1
• B. ≃ 0.6 × 10⁶ ms^–1
• C. ≃ 61 × 10³ ms^–1
• D. ≃ 0.3 × 10⁶ ms^–1

Answer 1

The Correct Option is A, B

To solve this problem, we will use the concept of the photoelectric effect and the associated equations.

The photoelectric effect is described by the equation:

K.E. = h\nu - \phi

where:

• K.E. is the kinetic energy of the ejected electron,
• h is Planck's constant,
• \nu is the frequency of the incident light,
• \phi is the work function of the metal, related to the threshold wavelength by \phi = \frac{hc}{\lambda_0}, where \lambda_0 is the threshold wavelength.

Given data:

• Threshold wavelength of silver \lambda_0 = 3250 \times 10^{-10} \ \text{m}
• Wavelength of ultraviolet light \lambda = 2536 \times 10^{-10} \ \text{m}
• Planck's constant h = 4.14 \times 10^{-15} \ \text{eVs}
• Speed of light c = 3 \times 10^{8} \ \text{ms}^{-1}

First, calculate the work function \phi using the threshold wavelength:

\phi = \frac{hc}{\lambda_0} = \frac{4.14 \times 10^{-15} \cdot 3 \times 10^{8}}{3250 \times 10^{-10}} \phi = \frac{12.42 \times 10^{-7}}{3250 \times 10^{-10}} = 3.821 \ \text{eV}

Now, calculate the energy of the incident ultraviolet light:

E_{\text{incident}} = \frac{hc}{\lambda} = \frac{4.14 \times 10^{-15} \cdot 3 \times 10^{8}}{2536 \times 10^{-10}} E_{\text{incident}} = \frac{12.42 \times 10^{-7}}{2536 \times 10^{-10}} = 4.898 \ \text{eV}

The kinetic energy of the ejected electron is:

K.E. = E_{\text{incident}} - \phi = 4.898 \ \text{eV} - 3.821 \ \text{eV} = 1.077 \ \text{eV}

Convert this kinetic energy to joules (1 eV = 1.6 \times 10^{-19} \ \text{J}):

K.E. = 1.077 \times 1.6 \times 10^{-19} \ \text{J} = 1.7232 \times 10^{-19} \ \text{J}

The kinetic energy is also related to velocity by the equation K.E. = \frac{1}{2}mv^2, where m is the mass of the electron (9.11 \times 10^{-31} \ \text{kg}).

Thus, solving for the velocity:

v = \sqrt{\frac{2 \cdot \text{K.E.}}{m}} = \sqrt{\frac{2 \cdot 1.7232 \times 10^{-19}}{9.11 \times 10^{-31}}} v = \sqrt{\frac{3.4464 \times 10^{-19}}{9.11 \times 10^{-31}}} = \sqrt{3.784 \times 10^{11}} v \approx 6.15 \times 10^{5} \ \text{ms}^{-1}

Therefore, the velocity of the electron ejected from the silver surface is approximately 6 \times 10^{5} \ \text{ms}^{-1} or 0.6 \times 10^{6} \ \text{ms}^{-1}, which matches the given option ≃ 6 \times 10^5 \ \text{ms}^{-1} or ≃ 0.6 \times 10^6 \ \text{ms}^{-1}.