Question

If the electron in hydrogen atom jumps from second Bohr orbit to ground state and difference between energies of the two states is radiated in the form of photons. If the work function of the material is $4.2eV$ then stopping potential is
[Energy of electron in $n^{th}orbit=-\frac{13.6}{n^2}eV]$

• A. $2eV$
• B. $4eV$
• C. $6eV$
• D. $8eV$

Answer 1

Answer: (C)

$6eV$

$E=-\frac{13.6}{4}+\frac{13.6}{1}=10.2$

So stopping potential equal to

$hv={\varphi }_0+V_s$

$10.2-4.2=V_s$

$V_s=6eV$