Question

In photo electric effect, if threshold wave length of a metal is \(5000 \AA\) work function of this metal is ..........V. $\left(\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}, 1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}\right)$ (A) \(1.24\) (B) \(2.48\) (C) \(4.96\) (D) \(3.72\)

Short answer

The work function of the metal is approximately \(2.49 \, eV\), which corresponds to option (B) \(2.48 \, eV\).

Step-by-step solution

Write down the threshold wavelength and the given constants

We are given the threshold wavelength λ, Planck's constant h, the speed of light c, and the conversion factor between electron volts and joules (1 eV = 1.6 x 10^-19 J). Let's note them. Threshold wavelength, λ = 5000 Ǻ Planck's constant, h = 6.62 x 10^-34 Js Speed of light, c = 3 x 10^8 m/s

Convert the threshold wavelength to meters

The threshold wavelength λ is given in angstroms (Ǻ). We need to convert it to meters (m) to match the SI units. 1 Ǻ = 10^-10 m So, λ = 5000 Ǻ x 10^-10 m/Ǻ = 5 x 10^-7 m

Use the photoelectric effect equation to find the energy of the photons

According to the photoelectric effect, the energy E of the photons in terms of the wavelength λ can be calculated by the following equation: E = h * (c / λ) Substitute the given values into the equation: E = (6.62 x 10^-34 Js) * (3 x 10^8 m/s) / (5 x 10^-7 m) E = 3.98 x 10^-19 J

Convert the energy to eV (electron volts)

Finally, we need to convert the energy from Joules (J) to electron volts (eV). Remember, 1eV = 1.6 x 10^-19 J. So, E (eV) = E (J) / (1.6 x 10^-19 J/eV) E (eV) = (3.98 x 10^-19 J) / (1.6 x 10^-19 J/eV) E (eV) ≈ 2.49 eV It is clear that the work function of this metal is approximately 2.49 eV, which corresponds to option (B) 2.48 eV.