Grade 12 Chemistry
A photon that has a frequency of 5.11 x 1014 Hz has an energy of_______________J of energy.
The options are:
a) 5.87 x 10^-8
b) 5.87 x 10^-6
c) 3.39 x 10^-11
d) 2.11 x 10^-5
e) 3.39 x 10^-18
We can use the formula E = hf, where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of the photon.
Substituting the given values:
E = (6.626 x 10^-34 J·s)(5.11 x 10^14 Hz)
E = 3.38186 x 10^-19 J
The energy of the photon is approximately 3.38 x 10^-19 J, which is closest to option e) 3.39 x 10^-18.
To find the energy of a photon with a given frequency, we can use the equation:
E = hf
Where:
E = Energy of the photon
h = Planck's constant (6.626 x 10^-34 J·s)
f = frequency of the photon
Let's substitute the given frequency into the equation and solve for E:
f = 5.11 x 10^14 Hz
h = 6.626 x 10^-34 J·s
E = (6.626 x 10^-34 J·s) × (5.11 x 10^14 Hz)
Multiplying the values together:
E = 3.37986 x 10^-19 J
Now we need to express this answer in scientific notation. We move the decimal point 19 places to the right:
E = 3.39 x 10^-19 J
Comparing this value with the options given, we can see that the closest one is:
e) 3.39 x 10^-18
Therefore, the correct answer is e) 3.39 x 10^-18 J.
To find the energy of a photon, you can use the equation:
Energy = Planck's constant (h) × frequency (ν)
where Planck's constant is approximately 6.626 × 10^-34 J·s.
Given the frequency of the photon as 5.11 × 1014 Hz, we can calculate its energy.
Energy = (6.626 × 10^-34 J·s) × (5.11 × 10^14 Hz)
Calculating this gives us:
Energy = 3.38 × 10^-19 J
Therefore, the correct option is e) 3.39 × 10^-18.