Question

(i) How does not explain the emission of electron from a photosensitive surface with the help of Einstein's photoelectric equation?
(ii) The work function of the following metals is given : $Na=2.75eV,K=2.3eV,Mo=4.17eV$ and $Ni=5.15eV$. Which of these metals will not cause photoelectric emission for radiation of wavelength $3300\stackrel{\circ }{A}$ from a laser source place $1m$ away from these metals? What happens if the laser source is brought nearer and place $50cm$ away?

Answer 1

Solution:-

i) In Photoelectric effect, an electron from of photosensitive surface, absorb a quantum of electromagnetic radiation($h\upsilon$). If this energy absorbed exceeds the minimum energy needed for the electron to escape from the photosensitive surface(work function${\varphi }_o$), the electrons emitted with maximum kinetic energy$\left(K_{max}\right)$.

$K_{max}=h\upsilon -{\varphi }_o$

This is Einstein's photoelectric equation.

ii) Given:-

${\left({\varphi }_o\right)}_{Na}=2.75eV{\left({\varphi }_o\right)}_{M_0}=4.17eV{\left({\varphi }_o\right)}_K=2.3eV{\left({\varphi }_o\right)}_{Ni}=5.15eV$

$\lambda =3300\mathring{A}$

Solution:-

energy of incident photon=$\frac{hc}{\lambda }=\frac{1.989\times {10}^{-25}}{3300\times {10}^{-10}}$

$=0.000603\times {10}^{-15}J$

$E=3.767eV$

$\therefore$ Since this energy is less than${\left({\varphi }_o\right)}_{M_0}\&{\left({\varphi }_o\right)}_{Ni}$

molybdenum & nickel will not exhibit photoelectric effect.

$\therefore$The distance of the laser is of no consequence.