Energy of photon. E=310(nm)1240=4eV>2eV (So photoelectric effect will take place)
=4×1.6×10−19=6.4×10−19joule
Number of photons falling per second
=6.4×10−196.4×10−5×1=1014
Number of photoelectron emitted per second
=1031014=1011
JEE Main 2020 — Physics Modern Physics
A beam of electromagnetic radiation of intensity 6.4×10−5W/cm2 is comprised of wavelength, λ=310nm . It falls normally on a metal (work function ϕ=2eV ) of surface area of 1cm2 . If one in 103 photons ejects an election, total number of electrons ejected in 1s is 10x . (hc=1240eVnm,1eV=1.6×10−19J), then x is ___________
Held on 7 Jan 2020 · Verified 6 Jul 2026.
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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 $\_\_\_\_$ $\Omega$. 
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The work function of a metal is 4.2 eV. The threshold wavelength for photoelectric emission is approximately:
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