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Electrons with a kinetic energy of $6.023 \times 10^4 \mathrm{~J} / \mathrm{mol}$ are evolved from the surface of a metal, when it is exposed to radiation of wavelength of $600 \mathrm{~nm}$. The minimum amount of energy required to remove an electron from the metal atom is
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Verified Answer
The correct answer is:
$2.3125 \times 10^{-19} \mathrm{~J}$
$\begin{aligned} & \because 1 \mathrm{~mol}=6.023 \times 10^{23} \text { atoms } \\ & \mathrm{KE} \text { of } 1 \mathrm{~mol}=6.023 \times 10^4 \mathrm{~J} \\ & \text { or KE of } 6.023 \times 10^{23} \text { atoms } \\ & =6.023 \times 10^4 \mathrm{~J} \\ & \therefore \quad \text { KE of } 1 \text { atom }=\frac{6.023 \times 10^4}{6.023 \times 10^{23}} \\ & =1.0 \times 10^{-19} \mathrm{~J} \\ & h v_{\text {energy }}=\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3 \times 10^8}{600 \times 10^{-9}} \\ & =3.313 \times 10^{-19} \mathrm{~J} \\ & \end{aligned}$
Minimum amount of energy required to remove an electron from the metal ion (ie, Threshold energy)
$\begin{aligned}
& =h v-\mathrm{KE} \\
& =3.313 \times 10^{-19}-1.0 \times 10^{-19} \\
& =2.313 \times 10^{-19} \mathrm{~J}
\end{aligned}$
Minimum amount of energy required to remove an electron from the metal ion (ie, Threshold energy)
$\begin{aligned}
& =h v-\mathrm{KE} \\
& =3.313 \times 10^{-19}-1.0 \times 10^{-19} \\
& =2.313 \times 10^{-19} \mathrm{~J}
\end{aligned}$
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