A 600 W mercury lamp emits monochromatic radiation of wavelength 331.3 nm. How many photons are emitted from the lamp per second?
h = 6.62 × 10⁻³⁴ Js velocity of light = 3 × 10⁸ ms⁻¹
(a) 1.0 × 10¹⁹
(b) 1.0 × 10²³
(c) 1.0 × 10²¹
(d) 2.0 × 10²³
Answers
Here f is the frequency, h is the planck's constant.
Or
Let n = number of photons/second
i.e
Solving for n:
This assumes that it converts all the power into light energy (though in reality it only converts about 5-9%, hence the answer would be 1/20th of this, but it's not in the options)
Answer:
Explanation:
Let's first write down our givens for this problem
Given
Power
=
600
W
or
600
J
s
λ
=
331.3
nm
or
(
3.313
⋅
10
−
7
m
)
Now let's try to establish a few things.
The mercury lamp is emitting a certain amount of
Energy
per
second
, defined as its
Power
. The source is coming from a monochromatic radiation with a wavelength of
3.313
⋅
10
−
7
m
. We are asked to find how many photons are being emitted from the lamp that is providing
600
J
of energy per second.
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
Step 1: Figure out the energy associated with the photon
We use the following formula
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
E
=
h
⋅
f
Where
E = Energy of the photon (J)
h = Planck's constant
(
6.62
⋅
10
−
34
J*s
)
f = frequency of the photon
(
1
s
)
But we aren't given the frequency; we are give the wavelength.
Well, we know that the speed of light is constant and given as
3.00
⋅
10
8
m
s
and can be calculated by using the following:
a
a
a
a
a
a
a
a
a
a
a
a
a
c
=
f
⋅
λ
Where
c
=
speed of light
(
3.00
⋅
10
8
m
s
)
f
=
frequency of radiation
(
1
s
)
λ
=
wavelength
(
m
)
Knowing this, we can isolate to solve for the frequency
a
a
a
a
a
a
a
a
a
a
a
a
a
c
=
f
⋅
λ
→
c
λ
=
f
We can replace
f
in the energy of the photon equation with
c
λ
and solve.
E
=
h
⋅
f
→
E
=
h
c
λ
E
=
(
6.62
⋅
10
−
34
J
⋅
s
)
(
3.00
⋅
10
8
m
s
)
3.313
⋅
10
−
7
m
→
5.99
⋅
10
−
19
J
Step 2: Use dimensional analysis to figure out photons emitted per s
Well, what does this tell us? This tells us that 1 photon has
5.99
⋅
10
−
19
J
a
a
a
a
a
a
a
a
a
a
a
a
a
a
1
photon
5.99
⋅
10
−
19
J
We were also told that the lamp produced
600
J
per second
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
600
J
s
We can use dimensional analysis to figure out the number of photons emitted per second.
a
a
a
a
a
600
J
s
⋅
1
photon
5.99
⋅
10
−
19
J
→
1.00
⋅
10
21
photons
s
Answer
:
1.00
⋅
10
21
photons
s