A metal surface having threshold frequency 2 x 1016 Hz is irradiated using photons of frequency V₁ = 11 x 1016 Hz and V₂ = 5 x 10¹6 Hz. The ratio of maximum kinetic energy of the emitted photoelectrons is
Answers
Answer:1. What is the energy in joules and electron volts of a photon of 420-nm violet light?
2. What is the maximum kinetic energy of electrons ejected from calcium by 420-nm violet light, given that the binding energy (or work function) of electrons for calcium metal is 2.71 eV?
Strategy
To solve Part 1, note that the energy of a photon is given by E = hf. For Part 2, once the energy of the photon is calculated, it is a straightforward application of KEe = hf − BE to find the ejected electron’s maximum kinetic energy, since BE is given.
Solution for Part 1
Photon energy is given by E = hf.
Since we are given the wavelength rather than the frequency, we solve the familiar relationship c = fλ for the frequency, yielding
f
=
c
λ
.
Combining these two equations gives the useful relationship
E
=
h
c
λ
.
Now substituting known values yields
E
=
(
6.63
×
10
−
34
J
⋅
s
)
(
3.00
×
10
8
m/s
)
420
×
10
−
9
m
=
4.74
×
10
−
19
J
Converting to eV, the energy of the photon is
E
=
(
4.47
×
10
−
19
J
)
1
eV
1.6
×
10
−
19
J
=
2.96
eV
Solution for Part 2
Finding the kinetic energy of the ejected electron is now a simple application of the equation KEe = hf − BE. Substituting the photon energy and binding energy yields KEe = hf − BE = 2.96 eV − 2.71 eV = 0.246 eV.
Explanation:
The ratio of maximum kinetic energy of the emitted photoelectrons is KE₁: KE₂ is 2:1
Explanation:
Kinetic energy is the energy which it posses during motion of the object.
Given frequency 2 x 10¹⁶ Hz
1) photons of frequency V₁ = 11 x 10¹⁶ Hz
KE₁=hv-hv°
= h (20×10¹⁶ × 11 x 10¹⁶)
=6.626×10⁻³⁴(2×10¹⁶ × 11 x 10¹⁶)
=1.45
frequency 2 x 10¹⁶ Hz
2)photons of frequency V₁ = 5x 10¹⁶ Hz
KE₂ =hv-hv°
= h (20×10¹⁶ × 11 x 10¹⁶)
=6.626×10⁻³⁴(2×10¹⁶ × 5 x 10¹⁶)
=0.662
ratio of KE₁: KE₂ is 2:1