Question

Calculate the threshold wavelength, in nm, of a metal if that metal emits an electron with a kinetic energy of 6.32 x 10-19 J when illuminated with light that has a frequency of 1.97 x 1015 Hz. 

Answer 1

$KE=\dfrac{hc}{\lambda}-W_o$

$W_o=h\nu-KE$

$W_o=(6.63\times10^{-34}\times1.97\times10^{15})-(6.32\times10^{-19})$

$W_o=6.74\times10^{-19}\space J$

For threshold wavelength$(\lambda)$ ,

$\dfrac{hc}{\lambda}=W_o$

$\lambda=\dfrac{hc}{W_o}=\dfrac{6.63\times10^{-34}\times3\times10^8}{6.74\times10^{-19}}$

$\lambda=2.95\times10^{-7}\space m$

$\lambda=295\space nm$