Question

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

Answer 1

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, $1eV=1.602\times {10}^{-19}J$
$\therefore 1eV\text{ato}{\text{m}}^{-1}=1.602\times {10}^{-19}J\text{ato}{\text{m}}^{-1}$
$\therefore 1eV\text{ato}{\text{m}}^{-1}=1.602\times {10}^{-19}\times 6.022\times {10}^{23}J\text{mo}{\text{l}}^{-1}$
$\therefore 1eV\text{ato}{\text{m}}^{-1}=96.45kJ\text{mo}{\text{l}}^{-1}$
$\therefore 1eV\text{ato}{\text{m}}^{-1}={\displaystyle \frac{96.45}{4.18}}Kcalmo{l}^{-1}$
$\therefore 1eV\text{ato}{\text{m}}^{-1}=23.06Kcal\text{mo}{\text{l}}^{-1}$
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 $KcalMo{l}^{-1}$ , 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 $KcalMo{l}^{-1}$ , divide by $4.18$ .