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Photoelectric Emission Dual Nature of Radiation and Matter Physics Class 12th

Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $^{}$ )

Option 1 -

$1.93 e V$

Option 2 -

$2.10 e V$

Option 3 -

$2.59 e V$

Option 4 -

$2.27 e V$

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1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$e V s = h v - ?$

$\begin{array}{r} At V_{s} = 0 \Rightarrow h v = ? \\ \Rightarrow ? = [6.62 \times 10^{- 34}] [10^{14}] [5.5] \\ \Rightarrow ? = \frac{ [6.62 \times 10^{- 34}] [10^{14}] [5.5] e V}{(1.6 \times 10^{- 19}]} \end{array}$

$= 2.27$

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Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $^{}$ )

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Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Plancks constant $^{}$ )