# Planck's problem (waves and frequency)

## Homework Statement

the frequency of A is doubled the threshold frequency of B.
if the wavelength of B is half that of A, what is the difference in speed?

E=hv
E=1/2mu^2
c=vλ

## The Attempt at a Solution

the answer is : square root of 3
i tried different ways,,,but just not getting anywhere.

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Simon Bridge
Homework Helper
It helps if you specify your problem completely.
I take it you mean this is a photoelectric effect problem?

light A has double the threshold frequency of light B (what determines the threshold frequency? Write it down.)

wavelength of B is half that of A (you can write that down?)

what is the difference in speed ... of what?
the two kinds of light? That would be zero.
the speed of electrons ejected by each type?
... do you have the relation between the energy of the incoming photon and the kinetic energy of the ejected electron?

"I tried different ways" does not tell us what you tried... we cannot help you if you don't tell us.

It helps if you specify your problem completely.
I take it you mean this is a photoelectric effect problem?

light A has double the threshold frequency of light B (what determines the threshold frequency? Write it down.)

wavelength of B is half that of A (you can write that down?)

what is the difference in speed ... of what?
the two kinds of light? That would be zero.
the speed of electrons ejected by each type?
... do you have the relation between the energy of the incoming photon and the kinetic energy of the ejected electron?

"I tried different ways" does not tell us what you tried... we cannot help you if you don't tell us.
OH YUP. sorry about that. this is the exact question.

Two photons (A&B) strike a metal surface and eject an electron.

The frequency of photon A is double the threshold frequency, while the wavelength of
photon B is ½ that of photon A.

What is the difference in speed (by what factor) of the electrons?

Express answer as a whole number.

my attempts:
i said the thresold frequency is = 1 hz
then A's frequency must be =2 hz
then the λ of A = (3.0x10^8m/s)/(2hz)=1.5x10^8m
B's λ is 1/2 of A = (1.5x10^8m/s)/2 = 0.75 x 10^8m
B's f is = (3.0x10^8m/s)/(0.75 x 10^8m) =4hz
i used the equation ΔEA=1/2mu^2 ΔEB=1/2mu^2 (i cancled m from noth equations)
ΔEA=1/2u^2=hv
$\sqrt{}hv/0.5$=u
for A = $\sqrt{}(6.62x10^-34)(2)/(0.5)$ = 5.14x10^-17m/s
i repeated the same thing for B (i used 4 hz for B) = i got 7.2x10^-17 m/s
its asking for the difference in speed
(7.2x10^-17)-(5.14x10^-17)=2.06x10^-17m/s
I'm no near the answer $\sqrt{}3$

Simon Bridge
Homework Helper
my attempts:
i said the thresold frequency is = 1 hz
why choose 1Hz? Why not just leave it as $\nu_{thresh}$ or $\nu_0$ or something?
then A's frequency must be =2 hz
then the λ of A = (3.0x10^8m/s)/(2hz)=1.5x10^8m
B's λ is 1/2 of A = (1.5x10^8m/s)/2 = 0.75 x 10^8m
B's f is = (3.0x10^8m/s)/(0.75 x 10^8m) =4hz
i used the equation ΔEA=1/2mu^2 ΔEB=1/2mu^2 (i cancled m from noth equations)
ΔEA=1/2u^2=hv
Doesn't this mean that all the energy of the incoming photon goes to the kinetic energy in the ejected electron?
Doesn't the work function have a say in this?