KEmax=hu−hu0hu−hu=e×ΔvV0=ehu−ehv0 ' v ' is doubled KEmax=2hu−hu0 V0′=(ΔV)′=e2hu−ehu0 KEmaxKEmax may not be equal to 2 ⇒V0V0′ may not equal to 2 KEmax=hu−hv0 V=ehv−eh h0
JEE Main 2011 — Physics Modern Physics
This question has Statement −1 and Statement −2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-1 : A metallic surface is irradiated by a monochromatic light of frequency v>v0 (the threshold frequency). The maximum kinetic energy and the stopping potential are Kmax and V0 respectively. If the frequency incident on the surface doubled, both the Kmax and V0 are also doubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
Held on 30 Apr 2011 · Verified 6 Jul 2026.
Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.
Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.
Statement-1 is false, Statement-2 is true.
Statement-1 is true, Statement-2 is false.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
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 $\_\_\_\_$ $\Omega$.
Work through every JEE Main Modern Physics PYQ, year by year.