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Standard XII

Chemistry

Nature of Electromagnetic Radiation

The maximum k...

Question

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6 e V$ fall on it is $4 e V$ . The stopping potential in volt is :

4

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Solution

The correct option is A $4$
Maximum Kinetic Energy, $K_{m a x} = \frac{h c}{λ} - φ$

$e V_{0} = K_{m a x}$

$e V_{0} = 4 e V$

$V_{0} = 4 V$

So, the answer is option (A).

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The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential in volt is :

The correct option is A 4Maximum Kinetic Energy, Kmax = hcλ−φ eV0 = Kmax eV0 = 4eV V0 = 4VSo, the answer is option (A).

The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy $6 e V$ fall on it is $4 e V$ . The stopping potential in volt is :

The correct option is A $4$
Maximum Kinetic Energy, $K_{m a x} = \frac{h c}{λ} - φ$

$e V_{0} = K_{m a x}$

$e V_{0} = 4 e V$

$V_{0} = 4 V$

So, the answer is option (A).