Question

When a radiation of wavelength \(3000 \AA\) is incident on metal, $1.85 \mathrm{~V}$ stopping potential is obtained. What will be threshed wave length of metal? $\left\\{\mathrm{h}=66 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(4539 \AA\) (B) \(3954 \AA\) (C) \(5439 \AA\) (D) \(4395 \AA\)

Short answer

The short answer is: The threshold wavelength of the metal is \(5439 \AA\) (Option C).

Step-by-step solution

Recall the photoelectric effect equation and Planck's equation

For the photoelectric effect, we know that the maximum kinetic energy of ejected electrons is given by: \(K_{max} = hf - hf_0\) where \(K_{max}\): maximum kinetic energy of ejected electrons \(h\): Planck's constant \(f\): frequency of incident radiation \(f_0\): threshold frequency The maximum kinetic energy of ejected electrons can also be expressed as: \(K_{max} = eV\) where \(e\): elementary charge \(V\): stopping potential Planck's equation relates the energy of a photon to its frequency: \(E = hf\) with \(E = \dfrac{hc}{\lambda}\) where \(\lambda\): wavelength of incident radiation \(c\): speed of light

Calculate the energy of the incident radiation

We are given the wavelength of the incident radiation, \(\lambda = 3000 \AA\), and we're supposed to find the energy of this radiation using the Planck's equation: \(E = \dfrac{hc}{\lambda}\) Plugging in the values of Planck's constant and speed of light, \(E = \dfrac{(6.6 \times 10^{-34} J.s)(3 \times 10^8 m/s)}{3000 \times 10^{-10} m}\) Calculate the value of energy: \(E = 6.6 \times 10^{-19} J\)

Calculate the maximum kinetic energy of ejected electrons

We're given the stopping potential, \(V = 1.85 V\), so we can calculate the maximum kinetic energy of ejected electrons using the relationship \(K_{max} = eV\). Since the elementary charge, \(e = 1.6 \times 10^{-19} C\), \(K_{max} = (1.6 \times 10^{-19} C)(1.85 V)\) Calculate the value of the maximum kinetic energy: \(K_{max} = 2.96 \times 10^{-19} J\)

Calculate the energy difference between the incident radiation and the ejected electrons

Use the photoelectric effect equation to find the energy difference: \(hf - hf_0 = K_{max}\) Since we know the incident energy from Step 2: \(hf - hf_0 = 6.6 \times 10^{-19} J - hf_0 = 2.96 \times 10^{-19} J\) Therefore, the threshold energy is \(hf_0 = 3.64 \times 10^{-19} J\)

Calculate the threshold wavelength

Now we have the threshold energy, and we can use the Planck's equation to calculate the threshold wavelength: \(\lambda_0 = \dfrac{hc}{hf_0}\) Plugging in the values for Planck's constant, speed of light, and the threshold energy: \(\lambda_0 = \dfrac{(6.6 \times 10^{-34} J.s)(3 \times 10^8 m/s)}{3.64 \times 10^{-19} J}\) Calculate the value of the threshold wavelength: \(\lambda_0 = 5.439 \times 10^{-7} m\)

Convert the threshold wavelength to angstrom and choose the correct option

Finally, convert the threshold wavelength from meters to angstrom: \(\lambda_0 = 5.439 \times 10^{-7} m \times \dfrac{10^{10} \AA}{1 m} = 5439 \AA\) So, the correct option is (C) \(5439 \AA\).