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The work function of the metal A is equ...

The work function of the metal A is equal to the ionization energy of the hydrogen atom in the first excited state. The work function of the metal B is equal to the ionization energy of `He^+` ion in the second orbit. Photons of the same energy E are incident on both A and B the maximum kinetic energy of photoelectrons emitted from A is twice that of photoelectrons emitted from B. Value of E (in eV) is

A

23.8 eV

B

20.8 eV

C

32.2 eV

D

24.6 eV

Text Solution

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The correct Answer is:
A
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Work function of metal X is equal to 3.5 eV and work function of material Y is equal to ionization energy of He+ ion in its first excited state. Light of same wavelength is incident on both X and Y. The maximum kinetic energy of photoelectrons emitted from X is twice that of photoelec- trons emitted from Y. Find the wavelength of incident light. hc = 12400 eV Å .

How many times is the ionization energy of He^(+) ion as compared to that of hydrogen atom ?

Knowledge Check

  • The ionization energy of hydrogen atom in the ground state is

    A
    13.6 MeV
    B
    13.6 eV
    C
    13.6 Joule
    D
    Zero

  • The ionization energy of Hydrogen atom in its ground state is……

    A
    3.4 e V
    B
    10.2 eV
    C
    13.6 eV
    D
    – 13.6 eV

  • Ionization energy of He^(+) ion at minimum energy position is

    A
    13.6 eV
    B
    27.2 eV
    C
    54.4 eV
    D
    68.0 eV

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