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  • In an experiment on photoelectric effect it was observed that for incident light of wavelength <img alt="1.98 \times 10^{-7}" src="https://entrancecorner.oncodecogs.com/gif.latex?1.98%20%5Ctimes%20

In an experiment on photoelectric effect it was observed that for incident light of wavelength 1.98 \times 10^{-7} , stopping potential is 2.5 \mathrm{V}. What is the work function \phi_{0} and threshold frequency?

Option: 1

3.75 \mathrm{eV}, 9.1 \times 10^{14} \mathrm{~Hz}


Option: 2

6.25 \mathrm{eV}, 9.1 \times 10^{14} \mathrm{~Hz}


Option: 3

6.25 \mathrm{eV}, 8.1 \times 10^{14} \mathrm{~Hz}


Option: 4

3.75 \mathrm{eV}, 8.1 \times 10^{14} \mathrm{~Hz}


Answers (1)

\begin{aligned} &\mathrm{K}_{\max }=\mathrm{eV}_{\mathrm{s}}=2.5 \mathrm{eV}\\ & \text{Where } \mathrm{V}_{\mathrm{s}} \text{ is the incident light}\\ & \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.98 \times 10^{-7}} \mathrm{~J} \\ & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.98 \times 10^{-7} \times 1.6 \times 10^{-19}} \mathrm{eV}=6.25 \mathrm{eV} \end{aligned}

According to Einstein's photoelectric equation

\begin{aligned} &\mathrm{K}_{\max }=\mathrm{E}-\phi_0 \text{ where }\phi_0 \text{ is the work function} \\ &\phi_0=\mathrm{E}-\mathrm{K}_{\max }=6.25 \mathrm{eV}-2.5 \mathrm{eV}=3.75 \mathrm{eV}\\ &\phi_0=\mathrm{hv}_0\\ &\text{ Where }\mathrm{v}_0 \text{ is the threshold frequency}\\ &\text{ or } \quad \mathrm{v}_0=\frac{\phi_0}{\mathrm{~h}}=\frac{3.7 \times 1.6 \times 10^{-19}}{6.6 \times 10^{-34}}=9.1 \times 10^4 \mathrm{~Hz} \end{aligned}

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Kshitij

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