100 HEURISTIC VIEW OF
LIGHT
energy
quanta
whose energy
is converted
at
least partially
to
kinetic
energy
of
electrons.
The simplest
possibility is that
a
light
quantum
transfers its
entire
energy
to
a
single
electron;
we
will
assume
that this
can occur.
However, we
will
not
exclude the possibility that the electrons absorb
only
a
part
of
the
energy
of
the
light
quanta.
An
electron
provided
with kinetic
energy
in the interior
of
the
body
will
have
lost
a
part
of
its kinetic
energy
[38]
by
the
time it reaches the surface. In addition, it will
have to be assumed
that in
leaving
the
body,
each
electron
has
to
do
some
work P
(characteristic
for the
body).
The
greatest
perpendicular velocity
on
leaving
the
body
will
be
that
of
electrons located directly
on
the surface
and
excited
perpendicular to it.
The
kinetic
energy
of
such
electrons is
J
ßv
-
P
.
If
the
body
is
charged
to
the positive
potential
II
and
is surrounded
by
conductors
of
zero
potential,
and
if
II
is
just
sufficient
to prevent
a
loss of electricity
of
the
body,
we
must
have
ne
=
$ ßv
-
P
,
[39]
where
e
denotes the electric
mass
of
the
electron,
or
HE
= Rßv
-
P]
,
where
E
denotes
the
charge
of
one
gram-equivalent
of
a
univalent
ion and
P1
is
the
potential of this quantity of
negative
electricity with
respect
to
the
body.1
[41]
If
one
sets
E
=
9.6
x
103,
then
II.10-8
is the
potential
in volts that
the
body
acquires
during
irradiation in the
vacuum.
To
see
whether
the relation derived
agrees
with
experience
in
order
of
magnitude,
we
put
P'
=
0,
v
=
1.03
x
1015
(which
corresponds to
the limit
1If
one assumes
that the release
of
the individual electron
from
a
neutral
molecule
by
light
must be accompanied
by
the
expenditure of
some
work,
one
does not have
to
change
anything
in the
above
relation;
but then P' is
to
be
considered
as
the
sum
of
two summands.
[40]