Rankers Physics

Photoelectric Effects and deBroglie Equation: Practice Problem & Solution

The work function of a surface of a photosensitive material is 6.2 eV. The wavelength of the incident radiation for which the stopping potential is 5 V lies in the: (2008)
X-ray region
Ultraviolet region
Visible region
Infrared region

Solution Explained:

To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:

The energy of incident radiation is $E = W + eV_0 = 6.2 eV + 5.0 eV = 11.2 eV$. The wavelength is $\lambda = \frac{1240 eV nm}{E(eV)} = \frac{1240}{11.2} \approx 110 nm$. This wavelength lies in the ultraviolet region of the electromagnetic spectrum.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. An electron in a hydrogen atom makes a transition from \(n = n_1\) to \(n = n_2\). The time period of revolution of the electron in the initial state is eight times that in final state. The possible value of \(n_1\) and \(n_2\) are
2. A mixture consists of two radioactive materials $A_1$ and $A_2$ with half lives of 20 s and 10 s respectively. Initially the mixture has 40 g of $A_1$ and 160 g of $A_2$. The amount of the two in the mixture will become equal after: (2012 Pre)
3. When an $\alpha$ particle of mass m moving with velocity v bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on mass: (2016 - I)
4. The mass number of He is 4 and that of sulphur is 32. The radius of sulphur nucleus is larger than that of helium by the factor of (1995)
5. When a hydrogen atom is raised from the ground state to an excited state, (1995)