The threshold frequency for a certain metal is `3.3xx10^(14) Hz.` If light of frequency `8.2xx10^(1)` Hz (hertz) is incident on the metal , find the cut - off voltage for photoelectric emission . Given Planck's constant `h = 6.625xx10^(-34)` Js (joule second ) . Charge of electron `e = 1.6 xx10^(-19)` C (coulomb).

Explanation

To find the cut-off voltage (stopping potential) for photoelectric emission, we can use the photoelectric effect equation derived from Einstein's theory:

\[ KE_{max} = h\nu - h\nu_0 \]

Where:
- \( KE_{max} \) is the maximum kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
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