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The value of $ 1eV{\text{ato}}{{\text{m}}^{ - 1}} $ is:
(A) $ 23.06{\text{ }}kcalmo{l^{ - 1}} $
(B) $ 96.45{\text{ }}kJmo{l^{ - 1}} $
(C) $ 1.602 \times {10^{ - 19}}{\text{ }}Jato{m^{ - 1}} $
(D) All of these

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Last updated date: 22nd Jul 2024
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Answer
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Hint : $ 1eV $ is defined as the energy gained by an electron when it has been accelerated by a potential difference of $ 1 $ volt. The work function of $ 1eV $ mainly depends on the properties of the metal and is highest for platinum with $ 5.65eV $ and lowest for caesium with $ 2.14eV $ .

Complete Step By Step Answer:
We know, $ 1eV = 1.602 \times {10^{ - 19}}J $
$\therefore 1eV{\text{ato}}{{\text{m}}^{ - 1}} = 1.602 \times {10^{ -19}}J{\text{ato}}{{\text{m}}^{ - 1}} $
$\therefore 1eV{\text{ato}}{{\text{m}}^{ - 1}} = 1.602 \times {10^{ - 19}} \times 6.022 \times {10^{23}}J{\text{mo}}{{\text{l}}^{ - 1}} $
$\therefore 1eV{\text{ato}}{{\text{m}}^{ - 1}} = 96.45kJ{\text{mo}}{{\text{l}}^{ - 1}} $
$\therefore 1eV{\text{ato}}{{\text{m}}^{ - 1}} = \dfrac{{96.45}}{{4.18}}Kcalmo{l^{ - 1}} $
$\therefore 1eV{\text{ato}}{{\text{m}}^{ - 1}} = 23.06Kcal{\text{mo}}{{\text{l}}^{ - 1}} $
Therefore, from the above expressions we can conclude that option (d) All of these is the correct answer.

Note :
To convert $ 1eVato{m^{ - 1}} $ to $ Kcal\;Mo{l^{ - 1}} $ , first convert it to $ Jmo{l^{ - 1}} $ using $ 1eV = 1.602 \times {10^{ - 19}}Jato{m^{ - 1}} $ and then multiply by Avogadro’s number $ 6.022 \times {10^{23}} $ . Now to convert the $ KJ $ into $ Kcal\;Mo{l^{ - 1}} $ , divide by $ 4.18 $ .