The wavelength of a particle having an Einste | vedclass

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(D) The Einstein energy $E$ is given by the formula: $E = \frac{N_A hc}{\lambda}$.
Given: $E = 1.995 	imes 10^{12} \ erg \ mol^{-1} = 1.995 	imes 1

Vedclass · Accepted Answer

$6000$

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(D) The Einstein energy $E$ is given by the formula: $E = \frac{N_A hc}{\lambda}$.
Given: $E = 1.995 	imes 10^{12} \ erg \ mol^{-1} = 1.995 	imes 10^5 \ J \ mol^{-1}$ (since $1 \ J = 10^7 \ erg$).
Using $N_A = 6.022 	imes 10^{23} \ mol^{-1}$,$h = 6.626 	imes 10^{-34} \ J \ s$,and $c = 3 	imes 10^8 \ m \ s^{-1}$:
$\lambda = \frac{N_A hc}{E} = \frac{6.022 	imes 10^{23} 	imes 6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{1.995 	imes 10^5}$.
$\lambda \approx \frac{1.197 	imes 10^{-1}}{1.995 	imes 10^5} \approx 6 	imes 10^{-7} \ m$.
Converting to $\mathring{A}: 6 	imes 10^{-7} \ m = 6000 \ \mathring{A}$.

The wavelength of a particle having an Einstein energy of $1.995 \times 10^{12} \ erg \ mol^{-1} \ photon^{-1}$ is .......... $\mathring{A}$.

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