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(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:$h = 6.62 	imes 10^{-34} \ J \cdot s$$c = 3 	imes 10^8 \ m/s$$\lam

Vedclass · Accepted Answer

$5.9 	imes 10^{-6} \ eV$

Vedclass · Answer

(B) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.Given:$h = 6.62 	imes 10^{-34} \ J \cdot s$$c = 3 	imes 10^8 \ m/s$$\lambda = 21 \ cm = 0.21 \ m$To convert the energy from Joules to electron-volts $(eV)$,we divide by $1.6 	imes 10^{-19} \ J/eV$.$E = \frac{6.62 	imes 10^{-34} 	imes 3 	imes 10^8}{0.21 	imes 1.6 	imes 10^{-19}} \ eV$$E = \frac{19.86 	imes 10^{-26}}{0.336 	imes 10^{-19}} \ eV$$E \approx 5.91 	imes 10^{-6} \ eV$.Therefore,the correct option is $B$.

An important spectral emission line has a wavelength of $21 \ cm$. The corresponding photon energy is $(h = 6.62 \times 10^{-34} \ J \cdot s; \ c = 3 \times 10^8 \ m/s)$.

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