The de Broglie wavelength of an electron is $0.4 � 10^{-10}$ m when its kinetic energy is $1.0\, keV$. Its wavelength will be $1.0 \times 10^{-10}\, m$, when its kinetic energy is
- 0.2 keV
- 0.8 keV
- 0.63 keV
- 0.16 keV
The Correct Option is D
Solution and Explanation
$\lambda=\frac{ h }{\sqrt{2 m E}}$
$ \therefore \frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{E_{2}}{E_{1}}} $
$\lambda_{1} =$ Wave length of electron
$=0 \cdot 4 \times 10^{-10} m $
$ E_{1} =1 \cdot 0\, keV $
$\lambda_{2} =1 \cdot 0 \times 10^{-10} m $
$E_{2} =? $
$\frac{0.4 \times 10^{-10}}{1 \times 10^{-10}} =\sqrt{\frac{E_{2}}{1 \cdot 0 keV }} $
$\frac{4}{10} =\sqrt{\frac{E_{2}}{1}} $
$\frac{16}{100} =\frac{E_{2}}{1} $
$ E_{2} =0 \cdot 16 \,keV$
Top Questions on de broglie hypothesis
If \( \lambda \) and \( K \) are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:
- JEE Main - 2025
- Physics
- de broglie hypothesis
- In the beginning of the 20th century, the matter wave concept was introduced by:
- OJEE - 2024
- Chemistry
- de broglie hypothesis
- The Principle of Uncertainty was introduced by:
- OJEE - 2024
- Chemistry
- de broglie hypothesis
- A proton and an electron have the same de Broglie wavelength. If \(K_p\) and \(K_e\) be the kinetic energies of proton and electron respectively. Then choose the correct relation:
- JEE Main - 2024
- Physics
- de broglie hypothesis
- A proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
Given: \(h = 6.63 \times 10^{-34} \, \text{Js}\), \(m_e = 9.0 \times 10^{-31} \, \text{kg}\), and \(m_p = 1836 \times m_e\)- JEE Main - 2024
- Physics
- de broglie hypothesis
Questions Asked in WBJEE exam
- Two straight conducting plates form an angle θ where their ends are joined. A conducting bar in contact with the plates and forming an isosceles triangle with them, starts at the vertex at time t = 0 and moves with constant velocity v to the right as shown in the figure. A magnetic field B points out of the page. The magnitude of emf induced at t = 1 second will be:
- WBJEE - 2024
- Electromagnetic induction
- A 2V cell is connected across the points A and B as shown in the figure. Assume that the resistance of each diode is zero in forward bias and infinity in reverse bias. The current supplied by the cell is:
- WBJEE - 2024
- Current electricity
- Consider a circuit where a cell of emf \( E_0 \) and internal resistance \( r \) is connected across the terminal A and B as shown in the figure. The value of \( R \) for which the power generated in the circuit is maximum, is given by:
- WBJEE - 2024
- Current electricity
- In a series LCR circuit, the rms voltage across the resistor and the capacitor are \(30 \, \text{V}\) and \(90 \, \text{V}\) respectively. If the applied voltage is \(50\sqrt{2}\sin\omega t\), then the peak voltage across the inductor is:
- WBJEE - 2024
- Alternating current
- Which of the following statement(s) is/are true in respect of nuclear binding energy?
(i) The mass energy of a nucleus is larger than the total mass energy of its individual protons and neutrons.
(ii) If a nucleus could be separated into its nucleons, an energy equal to the binding energy would have to be transferred to the particles during the separating process.
(iii) The binding energy is a measure of how well the nucleons in a nucleus are held together.
(iv) The nuclear fission is somehow related to acquiring higher binding energy.- WBJEE - 2024
- Nuclear physics
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.
Plank’s quantum theory relates the energy of an electromagnetic wave to its wavelength or frequency.
E = hν …….(1)
E = mc2……..(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 ‘v’ 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.