Question

Light of wavelength 600nm falls on a metal having photoelectric work function

2eV. Solve for (a) energy of photon (b) Kinetic energy of the most of energetic

photoelectron and (c) the stopping potential.​

Answer 1

Answer:

Chernozem is a black-colored soil containing a high percentage of humus and high percentages of phosphoric acids, phosphorus, and ammonia. Chernozem is very fertile and can produce high agricultural yields with its high moisture storage capacity.

Answer 2

Answer:

(a) $E = 33.1\times10^{-20}J = 2.06875 ev$

(b) K.E. =  0.6875 ev

(c) $V_o = 0.6875 V$

Explanation:

Given, wavelength (λ) = 600 nm

   and work function(Ф) = 2 eV

(a) Energy of photon(E) = $\frac{hc}{\lambda}$ (here $h = 6.603\times10^{-34} Js$, $c = 3\times10^{8} m/s$, and $\lambda = 600\times10^{-9}m$)

 $E = \frac{6.62\times10^{-34}\times3 \times10^{8}}{6\times10^{-7}}$

$E = 33.1\times10^{-20}J = 2.06875 ev$

(b) From photoelectric effect

       $K.E. = E - \phi$  (E = 2.06875 ev, Ф = 2 ev)

       K.E. =  0.6875 ev

(c) We know that stopping potential($V_{o}$) = $\frac{K.E.}{e}$

            $V_o = \frac{0.6875 ev}{e}$

   Therefore, $V_o = 0.6875 V$

#SPJ2