Maximum velocity of photoelectron emitted is 4-8 m-s- If e-m ratio of electron is 1-76 - 10 11 C kg-1- then stopping potential is given by -

The correct option is C 7× 10^ 11J/C eV_s=1/2mv^2_m V_s=mv^2_m/2e=v^2_m/2(e/m) =(4.8)^2/2× 1.76× 10^11 =7× 10^ 11J/C ...

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Standard XII

Physics

Stopping Potential

Maximum veloc...

Question

Maximum velocity of photoelectron emitted is $4.8 m / s$ . If e/m ratio of electron is $1.76 \times 10^{11} C k g^{- 1}$ , then stopping potential is given by :

5 \times 10^{- 10} J / C

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3 \times 10^{- 7} J / C

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7 \times 10^{- 11} J / C

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2.5 \times 10^{2} J / C

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Solution

The correct option is C $7 \times 10^{- 11} J / C$
$e V_{s} = \frac{1}{2} m v_{m}^{2}$
$V_{s} = \frac{m v_{m}^{2}}{2 e} = \frac{v_{m}^{2}}{2 (e / m)}$
$= \frac{(4.8)^{2}}{2 \times 1.76 \times 10^{11}}$

$= 7 \times 10^{- 11} J / C$

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Maximum velocity of photoelectron emitted is 4.8m/s. If e/m ratio of electron is 1.76×1011Ckg−1, then stopping potential is given by :

The correct option is C 7×10−11J/CeVs=12mv2mVs=mv2m2e=v2m2(e/m)=(4.8)22×1.76×1011=7×10−11J/C

Maximum velocity of photoelectron emitted is $4.8 m / s$ . If e/m ratio of electron is $1.76 \times 10^{11} C k g^{- 1}$ , then stopping potential is given by :

The correct option is C $7 \times 10^{- 11} J / C$
$e V_{s} = \frac{1}{2} m v_{m}^{2}$
$V_{s} = \frac{m v_{m}^{2}}{2 e} = \frac{v_{m}^{2}}{2 (e / m)}$
$= \frac{(4.8)^{2}}{2 \times 1.76 \times 10^{11}}$

$= 7 \times 10^{- 11} J / C$