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VITEEE 2009 Paper

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Question : 48 of 120

Marks: +1, -0

Electrons with a kinetic energy of

6.023 \times 10^{4} \mathrm{J/mol}

are evolved from the surface of a metal, when it is exposed to radiation of wavelength of 600 nm. The minimum amount of energy required to remove an electron from the metal atom is

$2.3125 \times 10^{-19} \mathrm{J}$
$3 \times 10^{-19} \mathrm{J}$
$6.02 \times 10^{-19} \mathrm{J}$
$6.62 \times 10^{-34} \mathrm{J}$

Solution:

∵

1 \mathrm{mol} = 6.023 \times 10^{23}

atoms

KE of

1 \mathrm{mol} = 6.023 \times 10^{4} \mathrm{J}

or KE of

6.023 \times 10^{23}

atoms =

6.023 \times 10^{4} \mathrm{J}

∴ KE of 1 atom =

\frac{6.023 \times 10^{4}}{6.023 \times 10^{23}} = 1.0 \times 10^{-19} \mathrm{J}

h v_{\text{energy}} = \frac{h c}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{600 \times 10^{-9}} = 3.313 \times 10^{-19} \mathrm{J}

Now since Threshold energy = hv - KE

= 3.313 \times 10^{-19} - 1.0 \times 10^{-19} = 2.313 \times 10^{-19} \mathrm{J}

Hence minimum amount of energy required to remove an electron from the metal ion will be

2.313 \times 10^{-19} \mathrm{J}.

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