Question

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

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