In a photo electric experiment 4 electrons with varying kinetic energy come out- E - 1-1-6- times 10-19 J CE - 2-2-1 10-19 J 1E - 3-3-2 10-19 J -E4 -1-2 10-19 J The stopping potential- given this data will be Volt

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Standard VIII

Biology

Method of finding LCM

In a photo el...

Question

In a photo electric experiment 4 electrons with varying kinetic energy come out.

$E_{1} = 1.6 \times 10^{- 19} J E_{2} = 2.1 \times 10^{- 19} J E_{3} = 3.2 \times 10^{- 19} J E_{4} = 1.2 \times 10^{- 19} J$

The stopping potential , given this data will be Volt

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Solution

Stopping potential is the potential required to stop the highest energy electrons from just hitting the collector.

The highest energy electron in the given setup is $E 3 = 3.2 \times 10^{- 19} J$

and energy is conserved , that we know for sure.

Initially the electron has kinetic energy = $3.2 \times 10^{- 19} J$

Finally it stops, so final K. E = 0

K.E. lost = $3.2 \times 10^{- 19} J$ …(i)

P.E. gained = $q Δ V$

= $- 1.6 \times 10^{- 19} (0 - V_{0})$

= $(1.6 \times 10^{- 19}) V_{0}$ …(ii)

$3.2 \times 10^{- 19} = 1.6 \times 10^{- 19} V_{0}$

$V_{0} = 2 v o l t .$

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In a photo electric experiment 4 electrons with varying kinetic energy come out.

$E 1 = 1.6 \times 10^{- 19} J E 2 = 2.1 \times 10^{- 19} J E 3 = 3.2 \times 10^{- 19} J E 1 = 1.2 \times 10^{- 19} J$

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In a photo electric experiment 4 electrons with varying kinetic energy come out.

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In a photo electric experiment 4 electrons with varying kinetic energy come out. E1=1.6×10−19JE2=2.1×10−19JE3=3.2×10−19JE4=1.2×10−19J The stopping potential , given this data will be Volt

In a photo electric experiment 4 electrons with varying kinetic energy come out.

$E_{1} = 1.6 \times 10^{- 19} J E_{2} = 2.1 \times 10^{- 19} J E_{3} = 3.2 \times 10^{- 19} J E_{4} = 1.2 \times 10^{- 19} J$

The stopping potential , given this data will be Volt