The work function of cesium is 2.14 eV The threshold frequency of caesium is, find the wavelength of the incident light if the stopping potential is 0.6 V

Question

The work function of cesium is 2.14 eV The threshold frequency of caesium is, find the wavelength of the incident light if the stopping potential is 0.6 V (A) 326 nm (B) 454 nm (C) 524 nm (D) 232 nm

Tardigrade Admin · Accepted Answer

According to Einsteinâs photoelectric equation eV0=h upsilon-φ0=(hc/λ)-φ0 (∵ upsilon=(c/λ)) or λ=(hc/(eV0+φ0))=(1242 eV nm/(0.6+2.14)eV) ≈454 nm

Q. The work function of cesium is $2.14 e V$ The threshold frequency of caesium is, find the wavelength of the incident light if the stopping potential is $0.6 V$

$326 nm$

$454 nm$

$524 nm$

$232 nm$

According to Einstein’s photoelectric equation
$e V_{0} = h υ - ϕ_{0} = \frac{h c}{λ} - ϕ_{0} (∵ υ = \frac{c}{λ})$
or $λ = \frac{h c}{( e V _{0} + ϕ _{0} )} = \frac{1242 e V nm}{( 0.6 + 2.14 ) e V} \approx 454 nm$

Q. The work function of cesium is 2.14eV The threshold frequency of caesium is, find the wavelength of the incident light if the stopping potential is 0.6V

326nm

454nm

524nm

232nm