White light (ranging in wavelengths from 380 to 750 nm) is incident on a metal w
ID: 1506126 • Letter: W
Question
White light (ranging in wavelengths from 380 to 750 nm) is incident on a metal with work function Wo = 2.62 eV.
a.What is the maximum kinetic energy of the electrons emitted from the surface?
KEmax =
b.For what range of wavelengths (from min to max) will NO electrons be emitted?
min = ? max =????
An electron in a hydrogen atom makes a transition from the n = 3 energy stage to the ground state (n=1). What is the frequency of the emitted photon?
f =
Now the electron is excited from its ground state to its first excited state. What is the frequency of the absorbed photon?
f =
While in this first excited state, the electron receives 1.889 eV of energy. What is the quantum number n that describes the energy level that the electron ends up at? (Enter an integer!)
n =
Explanation / Answer
A.
c = so = c/
max energy is from the shortest wavelength (highest frequency)
solve for = 3E8m/sec/380E-9m = 7.89E14Hz
Ephoton = h h = Planck's constant = 6.626068E-34 m² kg / s
E = 7.89E14 * 6.63E-34 = 5.23E-19J
E (in eV) = 5.23E-19/1.609E-19 = 3.24eV
Since energy is conserved, then it takes 2.62 eV of work to excite the metal so only
3.24 - 2.62eV of energy is left or 0.62eV
B.
electron emission will stop when the bombarding photons have energy less than the work function of 2.62eV so...
Set 2.62 eV = h and solve for
= 2.62(1.609E-19J)/6.63E-34 ...note how I switched from eV to Joules by multiplying by 1.609E-19
= 6.33E14 Hz
recall... c =
well, it follows that = c/
&lambda = 3E8 m/sec / 6.33E14 Hz = 473E-9m or 473nm and this equals lmin or the place where electrons leave with zero kinetic energy.
At any wavelengths less than 473nm, no electrons will be emitted at all.
lmax is quoted in the problem as 750nm.
So you can express this as:
No photon emission with between 499nm to 750nm.
---------------------------------------------------------------------------------------------------------------------------
a) Delta E = 13.6*( [1 / 1^2] - [1 / 3^2] ) eV = 12.08 eV
lambda = h*c / Delta E = 1240 eV*nm / Delta E = 1240 eV*nm / 12.08 eV
lambda = 102.5 nm
f = c / lambda = (2.99 x 10^17 nm/sec ) / 102.5nm = 2.914x 10^15 Hz ********************
(deep ultraviolet)
b) Delta E = 13.6*( [1 / 2^2] - [1 / 1^2] ) eV = 13.6 ( -0.75) eV = -10.2 eV
lambda = h*c / Delta E = 1240 eV*nm / Delta E = 1240 eV*nm / -10.2 eV
lambda = 122 nm
f = c / lambda = (2.99 x 10^17 nm/sec ) / 122 nm = 2.45 x 10^15 Hz *********************
(still ultraviolet in the Lyman series)
c) n=1: -13.6 eV
n=2: -3.40 eV
n=3: -1.51 eV
n=4: -0.850 eV
From part b) you are already at -3.40eV (n=2). Adding +1.888eV gets you to -1.51 eV which
is the n=3 level.
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