The electric field associated with a light wave is given by
\(E = E_0~ (\sin \omega_1 t)~ (\sin \omega_2 t)\).
This light wave falls on a metal having a threshold frequency,
\(\nu_0.\) The maximum kinetic energy of the emitted photoelectrons will be: (
\(h\) is Planck's constant)
| 1. |
Either \(\dfrac{h \omega_{1}}{2 \pi}\) or \(\dfrac{h \omega_{2}}{2 \pi}\) |
| 2. |
Either \(\left(\dfrac{h \omega_{1}}{2 \pi}-h \nu_{0}\right)\) or \(\left(\dfrac{h \omega}{2 \pi}-h \nu_{0}\right)\) |
| 3. |
\(\dfrac{h\left(\omega_{1}+\omega_{2}\right)}{2 \pi}-h \nu_{0}\) |
| 4. |
Both \(\dfrac{h\left(\omega_{1}+\omega_{2}\right)}{2 \pi}-h \nu_{0}\) and \(\dfrac{h\left |\omega_{1}-\omega_{2}\right|}{2 \pi}-h \nu_{0}\) |