JEE MAIN - Chemistry (2023 - 30th January Morning Shift - No. 22)
The energy of one mole of photons of radiation of frequency $$2 \times 10^{12} \mathrm{~Hz}$$ in $$\mathrm{J} ~\mathrm{mol}^{-1}$$ is ___________. (Nearest integer)
[Given : $$\mathrm{h}=6.626 \times 10^{-34} ~\mathrm{Js}$$
$$\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$$]
Answer
798
Explanation
$$\mathrm{E=nhv}$$
$$=(6.022\times10^{23})(6.626\times10^{-34})\times(2\times10^{12})$$
$$=798.03$$ J
$$\approx$$ $$798$$ J
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