Magnetic Effects & Magnetism

JEE Main level, single correct, medium

139 of 276

Question 139 of 276

A metal surface is illuminated by a radiation of wavelength 4500A∘4500A^{^{\circ}}. The ejected photo-electron enters a constant magnetic field of 2mT2mT making an angle of 90∘90^{\circ} with the magnetic field. If it starts revolving in a circular path of radius 2mm2mm, the work function of the metal is approximately

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Answer: (A) 1.36eV1.36eV

From the Einstein's photoelectric equation K=hcλ−ϕ⇒ϕ=hcλ−KK=\dfrac{hc}{\lambda }-\phi \Rightarrow \phi =\dfrac{hc}{\lambda }-K
Now R=mvBq⇒v=RBqm⇒K=R2B2q22mR=\dfrac{mv}{Bq}\Rightarrow v=\dfrac{RBq}{m}\Rightarrow K=\dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m}
Therefore, ϕ=hcλ−R2B2q22m=6.63×10−34×3×1084500×10−10−(2×10−3×2×10−3×1.6×10−19)22×9.1×10−31⇒ϕ=(4.42−2.25)×10−19J=2.271.6eV=1.36eV\phi =\dfrac{hc}{\lambda }-\dfrac{{R}^{2}{B}^{2}{q}^{2}}{2m}=\dfrac{6.63\times {10}^{-34}\times 3\times {10}^{8}}{4500\times {10}^{-10}}-\dfrac{{\left(2\times {10}^{-3}\times 2\times {10}^{-3}\times 1.6\times {10}^{-19}\right)}^{2}}{2\times 9.1\times {10}^{-31}} \Rightarrow \phi =\left(4.42-2.25\right)\times {10}^{-19}J=\dfrac{2.27}{1.6}eV=1.36eV

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