JEE Main 2022 · 24 Jun · Shift 2 · Chemistry Q62

← JEE Main 2022 · 24 Jun · Shift 2

ChemistrySingle correct+4 / -1Question 62 of 90

The energy of one mole of photons of radiation of wavelength 300 nm is

(Given : $h=6.63\times10^{-34}$ J s, $N_A=6.02\times10^{23}\ \mathrm{mol^{-1}}$ , $c=3\times10^8\ \mathrm{m\,s^{-1}}$ )

235\ \mathrm{kJ\,mol^{-1}}

325\ \mathrm{kJ\,mol^{-1}}

399\ \mathrm{kJ\,mol^{-1}}

435\ \mathrm{kJ\,mol^{-1}}

Answer and worked solution

Official answer: C

Energy per mole. $E=N_A\dfrac{hc}\lambda=\dfrac{6.02\times10^{23}\times6.63\times10^{-34}\times3\times10^8}{300\times10^{-9}}\approx3.99\times10^5$ J mol $^{-1}$ .

So about 399 kJ mol $^{-1}$ .

(C)

Topics:Photoelectric effect and spectra

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JEE Main 2022 · 24 Jun · Shift 2, Chemistry question 62