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If lambda0 is the Threshold wavelength ...

If ` lambda_0` is the Threshold wavelength of a metal for photo-electron emission . If the metal is exposed to the light of wavelength ` lambda` then the velocity of ejected electron will be ` sqrt ((2h)/m ( lambda_0 - lambda) K )`. The value of (K) is :

A

speed of light

B

1

C

`C/(lamda_0lamda)`

D

`1/(lamdalamda_0)`

Text Solution

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The correct Answer is:
C
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Knowledge Check

  • If lambda_0 is the Threshold wavelength of a metal for photoelencron emisssion . If the metal is exposed to the light of wavelngth lambda then the velocity of ejectred electgron will be sqrt ((2h)/m ( lambda_0 - lambda) K ) . The value fo (K) is :

    A
    `1`
    B
    `c/( lambda_0 lambda)`
    C
    `1/( lambda lambda_0)`
    D
    `(c. lambda)/( lambda_0)`

  • If I_(0) is the threshold wavelength for photoelectric emission, 1 the wavelength of light falling on the surface of a metal and m is the mass of the electron, then the velocity of ejected electron is given by

    A
    `[(2h)/(m)(lambda - lambda)]^(1//2)`
    B
    `[(2hc)/(m)(lambda_(0)-lambda)]^(1//2)`
    C
    `[(2hc)/(m){(lambda_(0)-lambda)/(lambda lambda_(0))}]^(1//2)`
    D
    `[(2hc)/(m){(1)(lambda_(0))-(1)/(lambda)}]^(1//2)`

  • If lambda_(0) is the threshold wavelnegth of a metal and lambda is the wavelength of the incident radiation, the maximum velocity of the ejected electron from the metal would be

    A
    `[ ( 2hc)/( m )((lambda_(0) - lambda)/( lambda_(0) lambda)) ]^(1//2)`
    B
    `[ ( 2hc)/( m )((lambda-lambda_(0) )/( lambdalambda_(0))) ]^(1//2`
    C
    `[ ( 2hc)/( m ) ( lambda - lambda_(0) ) ]^(1//2)`
    D
    `[ ( 2hc)/( m ) ( lambda_(0) - lambda) ]^(1//2)`

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