Question

Light of wavelength $180nm$ ejects photo-electrons from a plate of metal whose work-function is $2eV$. If a uniform magnetic field of $5\times {10}^{-5}$ T be applied parallel to the plate, what would be the radius of the path followed by electrons ejected normally from the plates with maximum energy. $(h=6.62\times {10}^{-34}$ joule-sec, $m=9.1\times {10}^{-31}Kg$ and $e=1.6\times {10}^{-19}$ coulomb).

Answer 1

$\text{We know,}$

$\frac{hc}{\lambda }-\varphi =K{E}_{max}$

$\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{180\times {10}^{-9}}-2\times 1.6\times {10}^{-19}=K{E}_{max}$

$\frac{19.86\times {10}^{-17}}{180}-3.2\times {10}^{-19}$

$11.03\times {10}^{-19}-3.2\times {10}^{-19}=7.83\times {10}^{-19}=K{E}_{max}$

$\text{Now,}$

$R=\frac{\sqrt{2\times m\times KE}}{qB}=\frac{\sqrt{2\times 9.1\times {10}^{-31}\times 7.83\times {10}^{-19}}}{1.6\times {10}^{-19}\times 5\times {10}^{-5}}$

$=\frac{\sqrt{142.506\times {10}^{-50}}}{8\times {10}^{-24}}=\frac{11.937\times {10}^{-1}}{8}$

$=1.4921\times {10}^{-1}=0.1492m$

$\text{Radius}=0.1492\text{metres}$