Question

The electric field associated with a light wave is given by
E=E0 sin [(1.57×107m-1)(x-ct)].
Find the stopping potential when this light is used in an experiment on photoelectric effect with the emitter having work function 1.9 eV.

Answer 1

The stopping-potential value is 1.205 V.

Explanation:

given data in the question

Electric field, $\text { Electric field, } E=E_{0} \sin \left[\left(1.57 \times 10^{7} \mathrm{m}^{-1}\right)(x-c t)\right]$

Work function, ϕ= 1.9 eV

Step 1:

When the given equation is compared with the standard equation,  

$E=E_{0} \sin (k x-w t)$ we get:

$\omega=1.57 \times 10^{7} \times c$    

frequency,            

$v=\frac{1.57 \times 10^{7} \times 3 \times 10^{8}}{2 \pi} H z$

Step 2:

From the Photoelectric Equation of Einstein,

$V_{0}$$=h v-\phi$

Here, $V_{0}$ = potential ends

          e = electron charge  

          h = Planck's constant

When the respective values are replaced we get:

$\begin{array}{c}{e V_{0}=6.63 \times 10^{-34} \times \frac{\left(1.57 \times 3 \times 10^{15}\right)}{\left(2 \pi \times 1.6 \times 10^{-19}\right)}-1.9 e V} \\{e V_{0}=3.105-1.9=1.205 e V}\end{array}$  

$V_{0}=\frac{1.205 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}}=1.205 V$

.

Answer 2

Answer:

stopping velocity is 1.605 v

hope it helps you