If light of frequency ` 8.2 xx 10^(14) Hz ` is incident on the metal , cut-off voltage for photoelectric emission is 2.03 V. Find the threshold frequency . Given ` h = 6.63 xx 10^(-34) Js , e = 1.6 xx 10^(-19)` C .
Text Solution
AI Generated Solution
To find the threshold frequency for the photoelectric emission, we can use Einstein's photoelectric equation:
\[ E = h \nu - \phi \]
Where:
- \( E \) is the kinetic energy of the emitted electrons.
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)).
- \( \nu \) is the frequency of the incident light.
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If light of frequency 8.2xx10^14 Hz is incident on the metal , cut-off voltage for photoelectric emission is 2.03 V.Find the threshold frequency . Given h= 6.63xx10^(-34) Js , e= 1.6xx10^(-19) C.
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