Q.1.A beam of electrons accelerated with 4.64 V was passed through a tube having mercury vapours .As a result of absorption,electronic changes occured with mercury atoms and light was emitted.If the full energy of the single electron ws converted into light ,what was the wave number of emitted light.Q.2.The binding energy of an electron in the ground state of an atom is equal to 24.6eV.Find the enegry required to remove both electrons from the atom.

Hi Vinay,

1) 4.64 eV is the energy of the electron

    if its total energy is converted photons then

    hc/λ=4.64*1.602*10^-19J

    1/λ=3.74*10⁶

2)B.E is 24.6 eV

   for hydrogen it is 13.6 eV

   24.6 which is nearly twice that of hydrogen implies the atom is He

   for first electron it is 24.6 eV

   for second electron it is 27.2 eV

   for removing both the electrons it is 27.2+24.6=51.8 eV

A.1)

V= 4.64V

E= QV

  = (1.6 x 10^-19 x 4.64) J

  = 7.424 x 10^-19 J

E=  hc /λ

  = hcv ,           [where v= 1/λ, v=wave number]

=> v= E/hc

=> v= (7.424 x 10^-19) / (6.62 x 10^-34 x 3 x 10⁸⁾ m^-1

=> v= 37400 cm^-1