Question

The threshold frequency of metal if the binding energy is 198.9 KJ mol${}^{-1}$ of electron is: [Given:$N_A=6\times {10}^{23}]$

• A. $5\times {10}^{14}{s}^{-1}$
• B. $6\times {10}^{14}{s}^{-1}$
• C. $7\times {10}^{14}{s}^{-1}$
• D. $8\times {10}^{14}{s}^{-1}$

Answer 1

Answer: (A)

$5\times {10}^{14}{s}^{-1}$

Binding energy = $198.9kJ/mol=\frac{198.9\times {10}^3}{6\times {10}^{23}}=3.32\times {10}^{-19}J/atom$
$E=h{\nu }_o=3.32\times {10}^{-19}J$
$\Rightarrow {\nu }_o=5\times {10}^{14}{s}^{-1}$