Photoelectric Effects and deBroglie Equation

Practice Questions:

Question 11: moderate

The anode voltage of a photocell is kept fixed. The wavelength λ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as given below :

View Answer

💬 Discuss this Question

Question 12: moderate

If the momentum of an electron is changed by P, then de-Broglie wavelength associated with it
changes by 0.2%. The initial momentum of electron will be about :

1. P/200

2. 100 P

3. 500 P

4. 400 P

View Answer

💬 Discuss this Question

Question 13: moderate

In photoelectric effect, the curve between photoelectric current and anode potential V (for
different frequencies) is shown in figure, then :

1. \[\nu_{1}>\nu_{2}>\nu_{3}\]

2. \[\nu_{1}<\nu_{2}<\nu_{3}\]

3. \[\nu_{1}=\nu_{2}=\nu_{3}\]

4. \[\nu_{1}>\nu_{2}<\nu_{3}\]

View Answer

💬 Discuss this Question

Question 14: moderate

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage
change in the de-Broglie wavelength of the particle is :

1. 25

2. 75

3. 60

4. 50

View Answer

💬 Discuss this Question

Question 15: moderate

The de Broglie wavelengths of a proton and an \(\alpha\)-particle are \(3\lambda\) and \(\lambda\) respectively. The ratio of the velocities of proton to \(\alpha\)-particle is

1. \(3 : 1\)

2. \(4 : 3\)

3. \(1 : 9\)

4. \(9 : 2\)

View Answer

Using the de Broglie wavelength formula \(\lambda = \frac{h}{mv}\), the velocity is (v = \frac{h}{m\lambda}\). Therefore, \(\frac{v_p}{v_\alpha} = \frac{m_\alpha \lambda_\alpha}{m_p \lambda_p} = \frac{4 m_p \cdot \lambda}{m_p \cdot 3\lambda} = \frac{4}{3}\).

💬 Discuss this Question

« Previous 1 2