If de-Broglie wavelength is \(\lambda\) when energy is E. Find the wavelength at \(\frac{E}{4}\) (Kinetic Energy).
- \(2\lambda \)
- \(\sqrt{2}\lambda \)
- \(\lambda \)
- \(\frac{\lambda }{\sqrt{2}}\)
The Correct Option is A
Solution and Explanation
\(\lambda=\frac{h}{p}\)
where h is Planck's constant and p is the momentum of the particle. The momentum of a particle with kinetic energy E is given by:
\(p = β(2mE)\)
where m is the mass of the particle.
If the de Broglie wavelength of a particle with energy E is π, then we have:
\(\lambda = \frac{h}{β(2mE)}\)
To find the de Broglie wavelength of a particle with kinetic energy E/4, we first need to find the momentum of the particle. The momentum is given by:
\(p = \sqrt{(2m(\frac{E}{4}))} = \sqrt{(m\frac{E}{2})}\)
The de Broglie wavelength of the particle with momentum p is then given by:
\(\lambda^{'} = \frac{h}{p} = \frac{h}{\sqrt(m\frac{E}{2})}\)
Dividing this expression by the original expression for π, we get:
\(\frac{{\lambda}'}{\lambda} = \sqrt{2}\)
So, the wavelength of the particle with kinetic energy \(\frac{E}{4}\) is β2 times the wavelength of the particle with kinetic energy E. Therefore, the answer is option 2: β2π.
Answer. A
Top Questions on Dual nature of radiation and matter
- A light wave described by E = 60[\(\sin(3 \times 10^{15})t + \sin(12 \times 10^{15})t\)] (in SI units) falls on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately)________ eV. (h=\(6.6 \times 10^{-34}\) J.s. and e=\(1.6 \times 10^{-19}\) C)
- JEE Main - 2026
- Physics
- Dual nature of radiation and matter
- What are de Broglie waves? Show that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).
- CBSE CLASS XII - 2026
- Physics
- Dual nature of radiation and matter
- The energy of an electron revolving in the nth orbit in Bohr model of hydrogen atom is proportional to:
- CBSE CLASS XII - 2026
- Physics
- Dual nature of radiation and matter
- The momentum of a photon associated with a microwave of wavelength 4.00 cm is:
- CBSE CLASS XII - 2026
- Physics
- Dual nature of radiation and matter
- Find the ratio of de-Broglie wavelengths of deuteron having energy E and \(\alpha\)-particle having energy 2E :
- CUET (UG) - 2026
- Physics
- Dual nature of radiation and matter
Questions Asked in JEE Main exam
The correct sequence of reagents for the above conversion of X to Y is :- JEE Main - 2026
- Periodic Table
- If \(x^2+x+1=0\), then the value of \(\left(x + \frac{1}{x}\right)^2 + \left(x^2 + \frac{1}{x^2}\right)^2 + \left(x^3 + \frac{1}{x^3}\right)^2 + \dots + \left(x^{25} + \frac{1}{x^{25}}\right)^2\) is:
- JEE Main - 2026
- Complex numbers
- Distance between an object and three times magnified real image is 40 cm. The focal length of the mirror used is ________ cm.
- JEE Main - 2026
- Wave optics
- For some \(\alpha, \beta \in R\), let \(A = \begin{pmatrix} \alpha & 2 \\1 & 2 \end{pmatrix}\) and \(B = \begin{pmatrix} 1 & 1 \\ \beta & 1 \end{pmatrix}\) be such that \(A^2-4A+2I-B^2-3B+I=O\). Then \((\det(\text{adj}(A^3-B^3)))^2\) is equal to:
- JEE Main - 2026
- Matrices and Determinants
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]- JEE Main - 2026
- Electrochemistry
Concepts Used:
De Broglie Hypothesis
One of the equations that are commonly used to define the wave properties of matter is the de Broglie equation. Basically, it describes the wave nature of the electron.
De Broglie Equation Derivation and de Broglie Wavelength
Very low mass particles moving at a speed less than that of light behave like a particle and waves. De Broglie derived an expression relating to the mass of such smaller particles and their wavelength.
Plankβs quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hΞ½ β¦β¦.(1)
E = mc2β¦β¦..(2)
As the smaller particle exhibits dual nature, and energy being the same, de Broglie equated both these relations for the particle moving with velocity βvβ as,
This equation relating the momentum of a particle with its wavelength is de Broglie equation and the wavelength calculated using this relation is the de Broglie wavelength.