Author Topic: Couple Q's (Read 1422 times) Tweet Share

sxcalexc · « **on:** August 25, 2008, 04:07:21 pm »

I'd love some help with these, preferably tonight, if possible

Collin Li · « **Reply #1 on:** August 25, 2008, 04:34:14 pm »

Question 11

$E_k_{(max)} = hf - W$

a) gradient: $h = 6.63 \times 10^{-34}$

b) Planck's constant - the energy associated with the frequency of light (since the units $\mbox{Js}$ can be rearranged to $\mbox{J/Hz}$ , and we use the formula $E_{photon} = hf$ to calculate the energy of a photon)

c) I think I am missing the "work function" of rubidium, but here's how you would do the question (just substitute in $W$ at the end)

$f = f_0$ (threshold frequency) when $E_k_{(max)} = 0$

$\implies f_0 = \frac{W}{h}$

Alternatively, read the "x-intercept" to find when $E_k_{(max)} = 0$ , hence grabbing the threshold frequency.

d) $f = \frac{3.00\times 10^{8}}{680 \times 10^{-9}} = 4.41 \times 10^{14} \mbox{ Hz}$

If $4.41 \times 10^{14} < f_0$ , then no. If $4.41 \times 10^{14} \geq 10^{14}$ , then yes.

sxcalexc · « **Reply #2 on:** August 25, 2008, 05:44:59 pm »

Holy crap soz Coblin I forgot you are given

How would you do C after having this table? Thx for your help, btw.

Collin Li · « **Reply #3 on:** August 25, 2008, 07:13:53 pm »

From the plot of your graph, using that data: read the "x-intercept" to find when E_k_{(max)} = 0, hence grabbing the threshold frequency.

orangez · « **Reply #4 on:** August 25, 2008, 08:01:47 pm »

Question 12

a) $W = hf_0$

$\Rightarrow W = \frac{6.63 \times 10^{-34}J s \times 5.01 \times 10^{14} Hz} {1.6 \times 10^{-19} C}$ the threshold frequency I used was attained from Q11c.

$\Rightarrow W = 2.07 V$

b) $E_k(max) = hf - W$

$\Rightarrow E_k(max) = 6.63 \times 10^{-34} Js \times 5.6 \times 10^{14} Hz - 2.07 eV \times 1.6 \times 10^{-19} C$

$\Rightarrow E_k(max) = 3.93 \times 10^{-20} J$

c) $E_k(max) = \frac{1}{2}mv^2$
rearrange the equation to find:
$v_{max} = 2.9 \times 10^5 m/s$

$p = mc$

$\Rightarrow p = 9.11 \times 10^{-31} kg \times 2.9 \times 10^5 m/s$ we use the speed of the least bound electron instead of the speed of light

$\Rightarrow p = 2.68 \times 10^{-25} kg m/s$

d) $E_k(max) = eV_0$

$\Rightarrow V_0 = \frac{3.93 \times 10^{-20} J}{1.6 \times 10^{-19}C}$

$\Rightarrow V_0 = 0.25 V$

sxcalexc · « **Reply #5 on:** August 25, 2008, 09:08:51 pm »

Are the answers wrong or something? :S How come it says threshold freq for rubidium is 4.4 and red light freq is 4.41 ... isn't that MORE than the threshold. I'm not getting 11 c still ...

PS. thx for reply orangez, where did you get your threshold freq value from? answer in the back says its 4.4 :S

Perhaps I'm missing an essential point here.

cara.mel · « **Reply #6 on:** August 25, 2008, 09:59:59 pm »

The answer they have given to 11c is wrong. The answer should be about 5*10^14 Hz. Thus, their answer to d is right

sxcalexc · « **Reply #7 on:** August 25, 2008, 10:05:58 pm »

Quote from: caramel on August 25, 2008, 09:59:59 pm

The answer they have given to 11c is wrong. The answer should be about 5*10^14 Hz. Thus, their answer to d is right

Ah, beautiful ... I was beginning to go a little crazy! That was the main problem I was having, haha. Nevertheless, thanks for your solutions, peeps!

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