If electromotive force of electron is $1.76 \times 10^{11} C K g^{- 1}$ and the stopping potential is $0.71 V$ , then the maximum velocity of the photo-electron is:-

150 k m s^{- 1}

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200 k m s^{- 1}

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500 k m s^{- 1}

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250 k m s^{- 1}

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Solution

The correct option is D $500 k m s^{- 1}$
Given that $\frac{q}{m} = 1.76 \times 10^{11} C / K g$
$V_{0} = 0.71 V o l t$
So max energy $E = q V$
and the maximum velocity will be given by $v = \sqrt[2]{\frac{2 E}{m}} = \sqrt[2]{\frac{2 q V_{0}}{m}} = \sqrt[2]{2 V_{0} \frac{q}{m}}$ .....(1)
putting corresponding values in the equation-1
we get $v = 4.999 \times 10^{5} m / s$ which nearly $500 k m / s$

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If electromotive force of electron is 1.76×1011CKg−1 and the stopping potential is 0.71V, then the maximum velocity of the photo-electron is:-

The correct option is D 500 km s−1Given that qm=1.76×1011C/KgV0=0.71VoltSo max energy E=qVand the maximum velocity will be given by v=2√2Em=2√2qV0m=2√2V0qm .....(1)putting corresponding values in the equation-1we get v=4.999×105m/s which nearly 500km/s

If electromotive force of electron is $1.76 \times 10^{11} C K g^{- 1}$ and the stopping potential is $0.71 V$ , then the maximum velocity of the photo-electron is:-