JEE MAIN - Chemistry (2022 - 28th June Morning Shift - No. 15)
If the work function of a metal is 6.63 $$\times$$ 10$$-$$19J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ____________ nm. (Nearest integer)
[Given : h = 6.63 $$\times$$ 10$$-$$34 J s, and c = 3 $$\times$$ 108 m s$$-$$1]
Answer
300
Explanation
Given,
Work function = 6.63 $$\times$$ 10$$-$$19 J
$$ = {{6.63 \times {{10}^{ - 19}}} \over {1.6 \times {{10}^{ - 19}}}}$$
= 4.14 eV
We know,
$$E = {{1240} \over {\lambda \,(nm)}}$$
$$ \Rightarrow 4.14 = {{1240} \over \lambda }$$
$$\lambda$$ = 300 nm
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