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Light of wavelength 0.6 mum from a sodiu...

Light of wavelength `0.6 mum `from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5 V. With wavelength `0.4 mum` from a sodium lamp, the stopping potential is 1.5 V. With this data , the value of h/e is

A

`4 xx 10^(-15) V`

B

`3 xx 10^(-15) V`

C

`4 xx 10^(-9) V`

D

`2 xx 10^(-9) V`

Text Solution

Verified by Experts

The correct Answer is:
1
`(w_(0))/(e)=(hc)/(lambda_(2)e)-V_(01),`
`(w_(0))/(e) = (hc)/(lambda_(2)e)-V_(02)`
`(h)/(c)c[(1)/(lambda_(1))-(1)/(lambda_(2))]=V_(01)-V_(02)`
`rArr(h)/(e)=(V_(01)-V_(02))/(c[(1)/(lambda_(1))-(1)/(lambda_(2))])`
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Knowledge Check

  • Light of wavelength 0.6 mum from a sodium lamp falls on a photocell and causes the emission of Photoelectrons for which the stopping potential is 0.5 V. With light of wavelength 0.4 mum from a sodium lamp, stopping potential is 1.5 V . With this data, the value of h/e is

    A
    ` 4 xx 10^(-19) V s `
    B
    `0.25 xx 10^(15) V s `
    C
    ` 4 xx 10^(-15) V s `
    D
    ` 4 xx 10^(-8) V s `

  • Light of wavelength 0.6mum from a sodium lamp falls on a photocell and causes the emission of photoelectrons for which the stopping potential is 0.5V. With light of wavelength 0.4mum from a murcury vapor lamp, the stopping potential is 1.5V . Then, the work function [in electron volts] of the photocell surface is

    A
    `0.75 eV`
    B
    `1.5eV`
    C
    `3eV`
    D
    `2.5eV`

  • In the question number 33, find the wavelength of the incident light if the stopping potential is 0.6 V.

    A
    326 nm
    B
    454 nm
    C
    524 nm
    D
    232 nm

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