Comparing Einstein's equation Kmax=hv−hv0, with y=mx+c, we get slope, m=h, which is Planck’s constant.
JEE Main 2024 — Physics Modern Physics
For the photoelectric effect, the maximum kinetic energy (Ek) of the photoelectrons is plotted against the frequency (v) of the incident photons as shown in figure. The slope of the graph gives

Held on 30 Jan 2024 · Verified 6 Jul 2026.
Ratio of Planck’s constant to electric charge
Work function of the metal
Charge of electron
Planck’s constant
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Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The electromagnetic wave exerts pressure on the surface on which they are allowed to fall. Reason (R): There is no mass associated with the electromagnetic waves. In the light of the above statements, choose the correct answer from the options given below :
The following diagram shows a Zener diode as a voltage regulator. The Zener diode is rated at $V_{z}=5 \mathrm{~V}$ and the desired current in load is 5 mA. The unregulated voltage source can supply upto 25 V. Considering the Zener diode can withstand four times of the load current, the value of resistor $R_{S}$ (shown in circuit) should be $\_\_\_\_$ $\Omega$. 
The minimum frequency of photon required to break a particle of mass 15.348 amu into $4 \alpha$ particles is $\_\_\_\_$ kHz. [mass of He nucleus = $4.002 \mathrm{amu}, 1 \mathrm{amu}=1.66 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J}. \mathrm{s}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ ]
The work function of a metal is 4.2 eV. The threshold wavelength for photoelectric emission is approximately:
A voltage regulating circuit consisting of Zener diode, having break-down voltage of 10 V and maximum power dissipation of 0.4 W, is operated at 15 V. The approximate value of protective resistance in this circuit is $\_\_\_\_$ $\Omega$.
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