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Answer 1 · 2017-12-02T14:58:15

(a) $3.37xx10^-19 (J)$
(b) $44.1(J)$
(c) $2.03xx10^5 (J)$

The energy of a photon is $E=hν$ , where
$h$ : Planck constant( $6.626×10^-34 J*s$ )
$ν$ : Frequency ( $ν=c/λ$ Hz)

The frequency of the light(λ= $589$ nm= $5.89×10^-7$ m) is
$ν= c/λ=(3.00xx10^8 " m/s")/(5.89xx10^-7 " m")=5.09xx10^14$ Hz.

(a) $E=hν=5.09xx10^14 Hz **6.626xx10^-34 J*s=3.37xx10^-19 (J)$ .

I will solve (c) first and get back to (b) later.

(c) Multiple the energy of a photon by Avogadoro's constant( $N_A=6.02xx10^23 mol^-1$ ).
$3.37xx10^-19*6.02xx10^23=2.03xx10^5 (J)$

(b) Molar mass of sodium is $23.0$ g/mol, and $5.00$ mg of $Na$ is equal to
$(5.00xx10^-3 g)/(23.0 g*mol^-1)=2.17xx10^-4$ mol.

Therefore, the total emitted energy is
$2.03xx10^5 (J*mol^-1) * 2.17xx10^-4 (mol) = 44.1 (J)$ .