Question 65 of 239

The radiation corresponding to $3\rightarrow 2$ transition of a hydrogen atom falls on a gold surface to generate photoelectrons. These electrons are passed through a magnetic field of $5\times {10}^{-4}T$. Assume that the radius of the largest circular path followed by these electrons is $7mm$, the work function of the metal is:
(Mass of electron $=9.1\times {10}^{-31}kg$)

A$1.36eV$B$1.88eV$C$0.16eV$D$0.82eV$

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Answer: (D) $0.82eV$

$3\rightarrow 2\Rightarrow 1.89eV$
$5\times {10}^{-4}Tr=7mm$
$r=\dfrac{mv}{qB}\Rightarrow mv=qrB$
$\Rightarrow E=\dfrac{{P}^{2}}{2m}=\dfrac{(qRB{)}^{2}}{2m}$
$=\dfrac{{\left(1.6\times {10}^{-19}\times 7\times {10}^{-3}\times 5\times {10}^{-4}\right)}^{2}}{2\times 9.1\times {10}^{-31}Joule}$
$=\dfrac{3136\times {10}^{-52}}{18.2\times {10}^{-31}\times 1.6\times {10}^{-19}}eV$
$=1.077eV$
We know work function $=$ energy incident $-$ $(KE{)}_{\text{electron}}$
$\phi =1.89-1.077=0.813eV$