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(D) According to the De Broglie relation,the momentum $p = \frac{h}{\lambda}$.Since $p = m 	imes c$,we have $m = \frac{h}{\lambda 	imes c}$.Given: $

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$3.746 	imes 10^{-36}$

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(D) According to the De Broglie relation,the momentum $p = \frac{h}{\lambda}$.Since $p = m 	imes c$,we have $m = \frac{h}{\lambda 	imes c}$.Given: $h = 6.6252 	imes 10^{-34} \, kg \cdot m^2/s$,$\lambda = 5894 \, \mathring{A} = 5894 	imes 10^{-10} \, m$,and $c = 3 	imes 10^8 \, m/s$.Substituting the values:$m = \frac{6.6252 	imes 10^{-34}}{5894 	imes 10^{-10} 	imes 3 	imes 10^8}$$m = \frac{6.6252 	imes 10^{-34}}{17682 	imes 10^{-2}}$$m = 0.00037468 	imes 10^{-32} \, kg$$m = 3.746 	imes 10^{-36} \, kg$.

If the wavelength is $5894 \, \mathring{A}$,the velocity of light is $3 \times 10^8 \, m/s$,and the value of $h$ is $6.6252 \times 10^{-34} \, kg \cdot m^2/s$,what will be the mass of a sodium photon?

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