Question

The radiation corresponding to $3\to 2$ 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. 1.36 eV
• B. 1.88 eV
• C. 0.16 eV
• D. 0.82 eV

Answer 1

Answer: (D)

$1.51-n=3$
$3.4-n=3$
$13.6-n=1$
$3\to 2\Rightarrow 1.89eV$
$5\times 10^{-4}Tr=7mm$
$r=\frac{mv}{qB}\Rightarrow mv=qrB$
$\Rightarrow E=\frac{P^2}{2m}=\frac{(qRB{)}^2}{2m}$
$=\frac{{\left(1.6\times 10^{-19}\times 7\times 10^{-3}\times 5\times 10^{-4}\right)}^2}{2\times 9.1\times 10^{-31}Joule}$
$=\frac{3136\times 10^{-52}}{18.2\times 10^{-31}\times 1.6\times 10^{-19}\times eV}$
$=1.077eV$
We know work function = energy incident $(KE{)}_{\text{electron }}$ $\varphi =1.89-1.077=0.813eV$