An electromagnetic radiation of wavelength $242 n m$ is just sufficient to ionise a sodium atom. Calculate the ionisation energy of sodium in kJ/mol?

Question

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Answer 1

From $E = h v$ or $h c / λ$
We have
$λ = 242$ nm
$= 242 \times 1 0^{- 9}$ m
$c = 3 \times 1 0^{8}$ m/s
$h = 6.62 \times 1 0^{- 34}$ Js

Now putting these values in the equation, we get

$E_{1} = \frac{( 6 . 6 2 \times 1 0 ^{- 34} J s \times 3 \times 1 0 ^{8} m / s )}{( 2 4 2 \times 1 0 ^{- 9} m )}$
$= 0.0821 \times 1 0^{- 17}$ J/atom

$E_{n} = E_{1} \times N_{A} \times \frac{1}{1 0 0 0} = \frac{( 0 . 0 8 2 \times 1 0 ^{- 17} ) \times 6 . 0 2 2 \times 1 0 ^{23}}{( 1 0 0 0 )}$
$= 494.1$ kJ/mol

Hence, option $A$ is correct .

Answer 2

From

E = h v

or

h c / λ

We have

λ = 242

nm

= 242 \times 1 0^{- 9}

m

c = 3 \times 1 0^{8}

m/s

h = 6.62 \times 1 0^{- 34}

Js

Now putting these values in the equation, we get

E_{1} = \frac{( 6 . 6 2 \times 1 0 ^{- 34} J s \times 3 \times 1 0 ^{8} m / s )}{( 2 4 2 \times 1 0 ^{- 9} m )}

= 0.0821 \times 1 0^{- 17}

J/atom

E_{n} = E_{1} \times N_{A} \times \frac{1}{1 0 0 0} = \frac{( 0 . 0 8 2 \times 1 0 ^{- 17} ) \times 6 . 0 2 2 \times 1 0 ^{23}}{( 1 0 0 0 )}

= 494.1

kJ/mol

Hence, option

A

is correct .