WB JEE 2024 | Dual Nature of Radiation Question 4 | Physics

A beam of light of wavelength $$\lambda$$ falls on a metal having work function $$\phi$$ placed in a magnetic field B. The most energetic electrons, perpendicular to the field are bent in circular arcs of radius R. If the experiment is performed for different values of $$\lambda$$, then $$\mathrm{B}^2$$ vs. $$\frac{1}{\lambda}$$ graph will look like (keeping all other quantities constant)

WB JEE 2024 Physics - Dual Nature of Radiation Question 4 English Option 1

WB JEE 2024 Physics - Dual Nature of Radiation Question 4 English Option 2

WB JEE 2024 Physics - Dual Nature of Radiation Question 4 English Option 3

WB JEE 2024 Physics - Dual Nature of Radiation Question 4 English Option 4

WB JEE 2017

MCQ (Single Correct Answer)

-0.25

When light of frequency v₁ is incident on a metal with work function W (where hv₁ > W), then photocurrent falls to zero at a stopping potential of V₁. If the frequency of light is increased to v₂, the stopping potential changes to V₂. Therefore, the charge of an electron is given by

$${{W({v_2} + {v_1})} \over {{v_1}{V_2} + {v_2}{V_1}}}$$

$${{W({v_2} + {v_1})} \over {{v_1}{V_1} + {v_2}{V_2}}}$$

$${{W({v_2} - {v_1})} \over {{v_1}{V_2} - {v_2}{V_1}}}$$

$${{W({v_2} - {v_1})} \over {{v_2}{V_2} - {v_1}{V_1}}}$$

WB JEE 2016

MCQ (Single Correct Answer)

-0.25

The potential difference V required for accelerating an electron to have the de-Broglie wavelength of 1 $$\mathop A\limits^o $$ is

100 V

125 V

150 V

200 V

WB JEE 2016

MCQ (Single Correct Answer)

-0.25

The work function of Cesium is 2.27 eV. The cut-off voltage which stops the emission of electrons from a cesium cathode irradiated with light of 600 nm wavelength is

0.5 V

$$-$$ 0.2 V

$$-$$ 0.5 V

No emission

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