Question

The magnetic field associated with a light wave is given, at the origin, by $B=B_0\left[sin\left(3.14\times {\left(10\right)}^7\right)ct+sin\left(6.28\times {\left(10\right)}^7\right)ct\right].$ If this light falls on a silver plate having a work function of $4.7eV,$ what will be the maximum kinetic energy of the photoelectrons?

$(c=3\times {\left(10\right)}^{8}m{s}^{-1},h=6.6\times {\left(10\right)}^{-34}Js)$

• A. $6.82eV$
• B. $7.72eV$
• C. $12.5eV$
• D. $8.52eV$

Answer 1

Answer: (B)

$B=B_0\left[sin\left(3.14\times {\left(10\right)}^{7}ct\right)+sin\left(6.28\times {\left(10\right)}^{7}ct\right)\right]$
Maximum $KE$ will associate with the maximum frequency
$f_{max}=\frac{6.28\times 10^{7}c}{2\pi }=10^7\times 3\times 10^{8}Hz=3\times 10^{15}Hz$
$K{E}_{max}=h{f}_{max}-\varphi$
$=\left[\frac{6.63\times 10^{-34}\times 3\times 10^{15}}{1.6\times 10^{-19}}-4.7eV\right]$
$=7.72eV$