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← JEE Main 2026 · 2 Apr · Shift 2
ChemistryInteger+4 / -1Question 71 of 75

JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 71

The surface of sodium metal is irradiated with radiation of wavelength xx nm. The kinetic energy of ejected electrons is 2.8×10−202.8\times10^{-20} J. The work function of sodium is 2.3 eV. The value of xx is ________ ×102\times10^2 nm. (Nearest integer)
(Given : h=6.6×10−34h=6.6\times10^{-34} J s ; 1 eV =1.6×10−19=1.6\times10^{-19} J; c=3.0×108c=3.0\times10^8 m s−1^{-1})

Answer and worked solution

Official answer: 5

Photon energy.
E=KE+ϕ=2.8×10−20+2.3×1.6×10−19=3.96×10−19 J.E=KE+\phi=2.8\times10^{-20}+2.3\times1.6\times10^{-19}=3.96\times10^{-19}\text{ J}.

Wavelength. λ=hcE=6.6×10−34×3×1083.96×10−19=5×10−7\lambda=\dfrac{hc}E=\dfrac{6.6\times10^{-34}\times3\times10^8}{3.96\times10^{-19}}=5\times10^{-7} m =5×102=5\times10^2 nm.

Answer: 5

Topics:Photoelectric effect and spectra
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