Question

An emission line of sodium has a wavelength of 330 nm. Calculate the energy of a photon of light emitted in J/ atom, and the energy emitted per mole of Na atoms at this wavelength.

Answer 1

Answer: The energy of the photon of light emitted is $6.024\times 10^{-19}J$ and energy emitted per mole of sodium atoms is $3.63\times 10^{5}J/mol$

Explanation:

The relation between energy and wavelength of light is given by Planck's equation, which is:

$E=\frac{hc}{\lambda}$

where,

E = energy of the photon

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3.0\times 10^8m/s$

$\lambda$ = wavelength of photon = 330 nm = $330\times 10^{-9}m$     (Conversion factor:  $1m=10^9nm$ )

Putting values in above equation, we get:

$E=\frac{6.626\times 10^{-34}Js\times 3.0\times 10^8m/s}{330\times 10^{-9}m}\\\\E=6.024\times 10^{-19}J$

To calculate the energy emitted per mole of sodium atom, we use the equation given by Planck's equation, which is:

$E=\frac{N_Ahc}{\lambda}$

where,

E = energy of the photon

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

h = Planck's constant = $6.626\times 10^{-34}Js$

c = speed of light = $3.0\times 10^8m/s$

$\lambda$ = wavelength of photon = 330 nm = $330\times 10^{-9}m$

Putting values in above equation, we get:

$E=\frac{6.022\times 10^{23}\times 6.626\times 10^{-34}Js\times 3.0\times 10^8m/s}{330\times 10^{-9}m}\\\\E=3.63\times 10^{5}J/mol$

Hence, the energy of the photon of light emitted is $6.024\times 10^{-19}J$ and energy emitted per mole of sodium atoms is $3.63\times 10^{5}J/mol$

Answer 2

The energy of the emission line is 9.97 × 10^ -42 J/ atom.

We have to recall that an atom emits energy when it moves form a higher to a lower energy level. This leads to the appearance of a line spectrum in the atom.

The energy emitted by sodium atom at a wavelength of 330nm can be obtained from;

E = hc/λ

h = Planks's constant

c = speed of light

λ = wavelength

Hence;

E = 6.6 × 10^-34 Js × 3  × 10^8 m/s / 330  × 10^-9 m

E = 6  × 10^-19 J

To obtain the unit in J/ atom;

6  × 10^-19 J/6.02 × 10^23 = 9.97 × 10^ -42 J/ atom

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