In a normal hydrogen atom,when the electron occupies its lowest available energy

Question

In a normal hydrogen atom,when the electron occupies its lowest available energy state the atom is said to be in its ground state.the maximum potential energy that an atom can have is 0J, at which point the electron has essentially been removed from the nucleus;thus the atom is ionized.how much energy will it take to ionize a hydrogen atom in its ground state?Calculate the wavelength of light that would be required to effect this ionization?Identify the series of spectral lines to which this wavelength belongs.

Explanation / Answer

ionization energy is given by:
En = -z2 13.6/ n2 ev
     = (-z2 13.6/n2 )* 1.6 x 10-19 J
where n is the state i.e for ground state its 1
and z is the atomic number and for hydrogen atom its 1

so En = -13.6* 1.6 * 10-19  J
         = -2.176 x10-18 J
so 2.176 x 10-18 J is required to remove theelectron from the ground state completely.
now we know that energy is given by:
hc/
where h is the plancks constant given by value: 6.63 x10-34 J-s
      c is the speed of the light givenby: 3 x 108 m/s
       and is thewavelength used
so here 2.176 x 10-18 = (6.63*10-34 )*(3*108 )/
= (6.63*10 -34 )*(3*108 )/ 2.176 x 10-18

= 9.1406 x 10-8 m
= 0.914 nm
so wavelenght of the light is = 0.914 nm
spectral lines to which the wavelength belongs is lymanseries i.e ultra violet region:

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