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The K.E. of electron is 10 keV. Calculat...

The K.E. of electron is 10 keV. Calculate its speed.

Text Solution

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`v=((2KE)/(m))^((1)/(2))" i.e. "v=((2xx10xx10^(3)xx1.6xx10^(-19))/(9.1xx10^(-31)))^((1)/(2))=((3.2)/(9.1))^((1)/(2))(10^(16))^((1)/(2))`
`v=0.5xx10^(8)ms^(-1)`.
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Knowledge Check

  • The K.E. of the electron is E, when the incident wavelength is lamda . To increase the K.E. of the electron to 2E, the incident wavelength must be :

    A
    `(hc)/(Elamda-hc)`
    B
    `(hclamda)/(Elamda+hc)`
    C
    `(hlamda)/(Elamda+hc)`
    D
    `(hclamda)/(Elamda-hc)`

  • If c is the speed of electromagnetic waves in vacuum, its speed in a medium of dielectric constant K and relative permeability mu_r is

    A
    `nu = 1/sqrt(mu_r K)`
    B
    `nu = c sqrt(mu_r K)`
    C
    `nu = c/sqrt(mu_r K)`
    D
    `nu = K/sqrt(mu_r c)`

  • The K.E. of the photoelectrons depends upon

    A
    intensity of radiation
    B
    frequency of radiation
    C
    the intensity and frequency of radiation
    D
    none of these

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