Question

Calculate the energy of one mole of photon of radiation whose frequency is 5* 10^14 Hz.

Answer 1

one mole of photon= 6.022*10^23 photons

we know

E=nhv

plank constant is 6.6* 10^-34

E= 6.022*10^23*6.6*10^-34*5*10^14

=198.7*10^3

1.99 * 10 approx

Answer 2

Answer:

$\large \implies199.62 \ kJ \ mol^{-1}$

Explanation:

Given :

$\displaystyle \text{Frequrency(v) = 5\times10^{14} \ Hz }$

We have to find energy ( e )

From planck's energy equation

we have

e = h v

where v = frequency

e = energy of photon

h = planck's constant

value of h = 6.63 × 10⁻³⁴ J sec

Putting values in equation we get

$\large e=6.63\times10^{-34}\times5\times10^{14}\\\\\\\large e=33.15\times10^{-20} \ J\\\\\\\large e=3.315\times10^{-19} \ J\\\\\\\large \text{Energy of one mole photons}\\\\\\\large \implies3.315\times10^{-19}\times6.022\times10^{23} \ J/mol\\\\\\\large \implies19.962\times10^{4} \ J \ mol^{-1}\\\\\\\large \implies199.62 \ kJ \ mol^{-1}$

Thus we get answer.