Skip to content
dronacharya.ai
← JEE Main 2014 · 6 Apr · Shift 1
PhysicsSingle correct+4 / -1Question 26 of 90

JEE Main 2014 · 6 Apr · Shift 1, Physics question 26

The radiation corresponding to 3→23\to2 transition of hydrogen atom falls on a metal surface to produce photoelectrons. These electrons are made to enter a magnetic field of 3×10−43\times10^{-4} T. If the radius of the largest circular path followed by these electrons is 10.0 mm, the work function of the metal is close to :

Choose one option
Answer and worked solution

Official answer: B

Photon. E=13.6(14−19)=1.89E=13.6\left(\frac14-\frac19\right)=1.89 eV.

Fastest electrons. p=eBr=1.6×10−19×3×10−4×10−2=4.8×10−25p=eBr=1.6\times10^{-19}\times3\times10^{-4}\times10^{-2}=4.8\times10^{-25} kg m/s, so
K=p22m=(4.8×10−25)22×9.1×10−31=1.27×10−19 J=0.79 eV.K=\frac{p^2}{2m}=\frac{(4.8\times10^{-25})^2}{2\times9.1\times10^{-31}}=1.27\times10^{-19}\text{ J}=0.79\text{ eV}.

Work function. ϕ=1.89−0.79≈1.1\phi=1.89-0.79\approx1.1 eV.

Topics:Photoelectric effectBohr model and hydrogen spectrumForce on a moving charge
← Previous questionNext question →