Question

How would I do this Physics problem?

Answer 1

Did your teacher tell you that
\[E _{photon} =\ h \nu\]and \[c = \lambda \nu\]?

Answer 2

the frequency corresponding to \lambda=435 nm, is:
\[\nu = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{4.35 \times {{10}^{11}}}}\]

Answer 3

oops..
\[\Large \nu = \frac{c}{\lambda } = \frac{{3 \times {{10}^8}}}{{4.35 \times {{10}^{ - 7}}}}\]

Answer 4

using your relation, which is the De Broglie formula, we get:
\[p = \frac{h}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}}}}{{4.35 \times {{10}^{ - 7}}}} = 1.52 \times {10^{ - 27}}\frac{{Kg \times m}}{{\sec }}\]

Answer 5

yes!

Answer 6

\[\Large p = \frac{h}{\lambda } = \frac{{6.63 \times {{10}^{ - 34}}}}{{4.35 \times {{10}^{ - 7}}}} = 1.52 \times {10^{ - 27}}\frac{{Kg \times m}}{{\sec }}\]

Answer 7

We can compute the kinetic energy of our electron, so we can write:
\[\Large KE = \frac{{{p^2}}}{{2m}} = \frac{{{{\left( {1.52 \times {{10}^{ - 27}}} \right)}^2}}}{{2 \times 9.11 \times {{10}^{ - 31}}}}\]

Answer 8

so your exercise is dealing with the photoelectric effect?

Answer 9

in order to apply the formula of the photoelectric effect, we have to know the so called work function of the material involved in that photoelectric experiment

Answer 10

please wait a moment, I'm trying to do some computations

Answer 11

I think that we have to use this formula:
\[\Large \frac{{{E_f} - {E_i}}}{h} = \frac{c}{\lambda }\]
since our electron makes a transition from an outer level of energy, Ei, to an inner level of energy Ef, inside the atom.
So our task is to pick a couple of levels, from your list, which gives a transition corresponding to your wave length of 435 nm

Answer 12

there E_i stands for initial energy level, whereas E_f stands for final energy level

Answer 13

we can rewrite the equation above, like below:
\[{E_f} - {E_i} = h\frac{c}{\lambda }\]
now, from my computation, I get:
\[h\frac{c}{\lambda } = 6.63 \times {10^{ - 34}}\frac{{3 \times {{10}^8}}}{{4.35 \times {{10}^{ - 7}}}} = 4.57 \times {10^{ - 19}}Joules\]
or:
\[\frac{{4.57 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 2.856\;eV\]

Answer 14

now if we pich the couple composed by the levels E2, and E5, and we compute the difference between the corresponding energies, we get:
\[{E_5} - {E_2} = - 0.544 + 3.403 = 2.859\;eV\]

Answer 15

which is equal to our previous result, so the transition made by our electron is between the levels E2 and E5 of your list

Answer 16

you have to divide like this:
\[\Large \frac{{4.57 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}} = 2.856\;eV\]

Answer 17

at moment we have used this formula:
\[\Large \Delta E = \frac{{hc}}{\lambda }\]
where \Delta E is the variation of energy of our electron during its atomic transition

Answer 18

so what we have to do is to searching for the right couple of levels which gives the right transition, namely that transition which corresponds to a wave length of 435 nm. In order to that we have to use the list of energy levels which your professor provided to us

Answer 19

your exercise is dealing with the atomic transition made by an electron