Question: Q. 12. A monochromatic light source of power $5 \mathrm{~mW}$ emits $8 \times 10^{15}$ photons per second. This light ejects photoelectrons from a metal surface. The stopping potential for this set up is $2 \mathrm{~V}$. Calculate the work function of the metal.
A [Delhi I, II, III 2016]
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Solution:
Ans.
$$ \begin{aligned} P & =5 \times 10^{-3} \mathrm{~W} \ n & =\frac{P}{E} \end{aligned} $$
$$ E=\frac{P}{n}=6.25 \times 10^{-19} \mathrm{~J} \quad 1 / 2 $$
$$ \begin{array}{rlrl} E & =3.9 \mathrm{eV} & 1 / 2 \ W_{0} & =E-\mathrm{eV}_{0} & 1 / 2 \end{array} $$
$$ =(3.9-2) \mathrm{eV} $$
$W_{0}=1.9 \mathrm{eV}$
Detailed Answer :
[CBSE Marking Scheme 2016]
Power of monochromatic light source
= total energy released per second
$=$ number of photons emits per second
$\times$ energy of one photon
Number of photons emits per second $=8 \times 10^{15}$
Power of monochromatic light source $=5 \mathrm{~mW}$
Hence, energy of one photon,
$$ \begin{aligned} h v & =\frac{5 \times 10^{-3}}{8 \times 10^{15}} \ & =0.625 \times 10^{-18} \mathrm{~J} \ & =\frac{0.625 \times 10^{-18}}{1.6 \times 10^{-19}}=3.9 \mathrm{eV} \end{aligned} $$
$K . E_{\text {max }}=e V_{0}$; where $V_{0}$ is stopping potential.
$$ \begin{aligned} & =V_{0} e V=2 e V\left(V_{0}=2 \text { Volts }\right) \ K . E . & =h v-\phi_{0} \ 2 & =3.9-\phi_{2} \ \phi_{0} & =(3.9-2) \mathrm{eV} \ \phi_{0} & =1.9 \mathrm{eV} \end{aligned} $$
Answering Tips
- First calculate the energy of one photon and convert it to $\mathrm{eV}$.
