If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
The correct option is C be more than double
According to Einstein's photoelectric equation, K.Emax=hν−W, which tells us the value of the maximum kinetic energy corresponding to the fastest electron in the photocurrent. This fastest electron with the maximum kinetic energy gets just stopped when the potential is kept at the stopping potential V0, i.e., the electron almost makes it to the collector plate before returning back to the emitter plate. Thus there is a point right near the collector plate where the whole of the kinetic energy, K.Emax gets converted into potential energy due to the potential V0. Thus, we can write -
eV0=K.Emax=hν−W
⇒v=ehV0+W
Let ν'=2ν, the doubled frequency, for which the new stopping potential is, say, . We want to find the relationship between V′0 and V0. Following the equation above, we can write , for ν' -
v′=ehV′0+W .....(1)
But also,
v′=2v=eh(2V0)+2W=eh(2V0+hWe)+W ....(2)
Comparing the equations (1) and (2),
⇒V′0=2V0+hWe>2V0
As you can see, option (c) is the correct choice!
According to Einstein's photoelectric equation, K.Emax=hν−W, which tells us the value of the maximum kinetic energy corresponding to the fastest electron in the photocurrent. This fastest electron with the maximum kinetic energy gets just stopped when the potential is kept at the stopping potential V0, i.e., the electron almost makes it to the collector plate before returning back to the emitter plate. Thus there is a point right near the collector plate where the whole of the kinetic energy, K.Emax gets converted into potential energy due to the potential V0. Thus, we can write -
eV0=K.Emax=hν−W
⇒v=ehV0+W
Let ν'=2ν, the doubled frequency, for which the new stopping potential is, say, . We want to find the relationship between V′0 and V0. Following the equation above, we can write , for ν' -
v′=ehV′0+W .....(1)
But also,
v′=2v=eh(2V0)+2W=eh(2V0+hWe)+W ....(2)
Comparing the equations (1) and (2),
⇒V′0=2V0+hWe>2V0
As you can see, option (c) is the correct choice!
