IIT JEE , AIEEE Forum - Physics , Modern Physics

raman_shadow

the electric field associated with a light wave is given as
E=E₀sin[(1.57x10⁷ m^-1)(ct-x)]

Find the stopping potential when this light is used in an experiment on photoelectric effect with emitter having work function 2.1eV.

edison

The stopping potential is related to the (maximum) kinetic energy of the ejected electrons by

eV₀ = KE_max = $\displaystyle{\textstyle\frac{1}{2}}$ mv_max².

Further,

KE_max = eV₀ = hf - $\displaystyle\phi$ = $\displaystyle{\frac{hc}{\lambda}}$ - $\displaystyle\phi$ .

Here E = Eo sin [1.57*10⁷(ct -x )]

so,

= 2

= 1.57*10⁷c

Now find

and substitute the value to find Stopping potential V_o

raman_shadow

please can u show me its full solution I require it urgently.............

thank you

malay

the general equation of a wave is
Asin(t-)
now concerntrate on the angle term
=2pi/T=2pi*c/lambda

=x/lambda

malay

picture is not working in last post
he general equation of a wave is
Asin(omega*t-theta)
omega=2pi/T=2pi*c/lambda
theta=x/lambda