Question

In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident light of wavelength 278 nm is used and 2.6 V is needed for light of wavelength 207 nm. From these data determine Planck's constant and the work function of the metal.

Answer 1

Answer:

Planck’s constant is $6.9*10^{-34} Js$

work function of metal is $5.846*10^{-19}$

Explanation:

The maximum kinetic energy from electron from photoelectric effect is given by

$\K_{max}=eV_o$ where $V_o$  is applied voltage and e is charge on electron and substituting charge on electron by $1.6*10^{-19}$  and 1.00 V for $V_o$  

$K_{max}=1.6*10^{-19}*1=1.6*10^{-19} J$

Considering that the wavelength, \lambda of light used is given as $278*10^{-9} m$

Energy of light is given by

$E=\frac {hc}{ \lambda}$   where $\lambda$  is the wavelength, h is Planck’s constant and c is speed of light. Taking c for $3*10^{8} m/s$  

Substituting values of wavelength and speed of light we obtain

$E=\frac {h*3*10^{8}}{278*10^{-9}}=h1.07914*10^{15} J$

If work function is $\phi$  then

$E=\phi + k_{max}$  hence

$h1.07914*10^{15} J=\phi +1.6*10^{-19} J$ ------- Equation 1

Energy corresponding to wavelength of 207 nm is

$E=h\frac {3*10^{8} m/s}{207*10^{-9}}=h(1.45*10^{15}}) J$

Maximum kinetic energy of electrons when $V_o$  is 2.6 V becomes

$K_{max}=(1.6*10^{-19})*2.6V=4.16*10^{-19} J$

From $E=\phi + k_{max}$  and substituting $h(1.45*10^{15}}) J$  for E and $4.16*10^{-19} J$  for $K_{max}$  we have

$h(1.45*10^{15}}) J=\phi +4.16*10^{-19} J$ ------ Equation 2

Equation 2 minus equation 1 gives

$h3.71*10^{14}=2.56*10^{-19}$

$h=\frac {2.56*10^{-19}}{3.71*10^{14}}\approx 6.9*10^{-34} Js$

Therefore, Planck’s constant is $6.9*10^{-34} Js$

Recalling equation 1 and substituting back the value of h as obtained

$h1.07914*10^{15} J=\phi +1.6*10^{-19} J$

$(6.9*10^{-34})*(1.07914*10^{15})=\phi +1.6*10^{-19} J$

$\ph=(6.9*10^{-34})*(1.07914*10^{15})-(1.6*10^{-19})=5.846*10^{-19}$

Therefore, work function of metal is $5.846*10^{-19}$