Question

For silver metal, the threshold frequency $v_o$ is $1.12\times {10}^{17}\text{Hz}$. What is the maximum kinetic energy of photoelectrons produced by shining U. V. light of $15.0\stackrel{o}{\text{A}}$ wavelength on the metal?

• A. $3.00\times {10}^{-17}\text{J}$
• B. $5.8\times {10}^{-17}\text{J}$
• C. $3.9\times {10}^{-17}\text{J}$
• D. $2\times {10}^{-15}\text{J}$

Answer 1

The correct option is B $5.8\times {10}^{-17}\text{J}$

We know that, $v=\frac{c}{\lambda }=\frac{3\times {10}^8{\text{m s}}^{-1}}{15\times {10}^{-10}\text{m}}=2\times {10}^{17}{\text{s}}^{-1}$
Applying Einstein's Photoelectric Equation,
$K.E.=hv-h{v}_o=h(v-v_o)$
$=6.626\times {10}^{-34}(2\times {10}^{17}-1.12\times {10}^{17})$
$=5.8\times {10}^{-17}\text{J}$