In this illustration, we'll be analyzing a
situation, where photo electrons incident,
on a hydrogen gas. we are given that a monochromatic
light source of frequency nu illuminates a
metallic surface and ejects photoelectrons.
and these photoelectrons are just able to
ionize hydrogen atom in ground state. when
fall on these. and when the experiment is
repeated with incident frequency changed to
5 nu by 6. it is asking the photoelectron
emitted are able to excite the hydrogen atom
which then emit, the radiation of wavelength
1 2 1 5 angstrom. so we are required to find
the work function of metal and the frequency.
so here, we can directly write. initially.
we use. h nu is equal to phi plus 13 point
6 electron volt. because, in this situation
we are given that photo electrons are just
able to ionize hydrogen atom. that means the
kinetic energy maximum kinetic energy of these
photoelectrons must be 13 point 6. as, we
are given that k e max is equal to ionization
energy that is 13 point 6 electron volt. for
the ejected electrons. and later we can say,
when the frequency is increased, the electrons
are able to excite the hydrogen atoms due
to which these atoms will emit the radiation
of wavelength 1 2 1 5 angstrom. so here we
can write when, nu is changed to, 5 nu by
6. h atom. emit. photon of energy. which is
given by 1 2 4 3 1, divided by 1 2 1 5. which
is 10 point 2 electron volt. and we know well
for hydrogen atom when. 10 point 2 electron
volt photon is emitted that means it is corresponding.
to transition. from n equal to 2 to, n equal
to, 1. so this is the transition which take
place when, hydrogen atoms are excited by
the photoelectrons that means, for transition
2 to 1 first electros, of hydrogen atom must
be excited from 1 to 2. so, this implies the
maximum kinetic energy. of photoelectrons.
is second case. must be. 10 point 2 electron
volt. so that hydrogen atom will excite from
1 to 2 and then it emits this radiation. this
implies here we use. h multiplied by 5 nu
by 6 that is equal to 5. plus 10 point 2 electron
volt. now in this situation say if we consider
this as equation 1 and this as equation 2.
then, subtracting these 2 values it gives,
if we, use 1 minus 2. this implies, h multiplied
by nu by 6. is equal to 3 point 4 electron
volt. and, on simplifying this gives us the
value of nu, that is 6 multiplied by 3 point
4 into 1 point 6 into 10 to power minus 19.
divided by the value of h is 6 point 6 3 into
10 to power minus 34. and this becomes 4 point
9 3 multiply by 10 to power 15 hertz. that
is 1 result of this problem. that is the frequency
nu. if we wish to calculate the work function,
then we can use from equation 1, we have phi
is equal to h nu minus, 13 point 6. and on
substituting the value. here, you can see
the value of h nu, we are getting is equal
to 6 multiplied by, 3 point 4 electron volt
that is 20 point 4. so here h nu is 6 multiplied
by 3 point 4 electron volt. minus 13 point
6, which is equal to 6 point 8 electron volt
that is another answer for this problem.
