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In a photo-emissive cell, with exciting ...

In a photo-emissive cell, with exciting wavelength `lambda`, the maximum kinetic energy of electron is K. If the exciting wavelength is changed to `( 3 lambda)/( 4)` the kinetic energy of the fastest emitted electron will be `:`

A

`3K //4`

B

`4K //3`

C

less than `4K//3`

D

greater than `4K//3`

Text Solution

Verified by Experts

The correct Answer is:
D
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Knowledge Check

  • In a photo-emissive cell, with exciting wavelength lambda_1 , the maximum kinetic energy of electron is K. If the exciting wavelength is changed to lambda_2 , the maximum kinetic energy of electron is 2K then

    A
    `lambda_1=2lambda_2`
    B
    `lambda_1 gt 2lambda_2`
    C
    `lambda_1 lt 2lambda_2`
    D
    `lambda_1=lambda_2/2`

  • In a photo emissive cell, with exciting wavelength X the maximum kinetic energy of electron is K. If the exciting wavelength is changed to 31/4 , then the kinetic energy of the fastest emitted electron will be

    A
    `(3K)/4`
    B
    `(4K)/3`
    C
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    D
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  • In a photoemissive cell, with exciting wavelength lambda , the faster electron has speed v. If the exciting wavelength is changed to 3lambda//4 , the speed of the fastest electron will be

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    B
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    C
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    D
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