The green light emitted by a stoplight has a wavelength of 525 nm. What is the frequency of this photon? (c = 3.00 × 10⁸ m/s).
1 answer:
The frequency of this photon is 5.94 x .
Solution:
The speed of light has a value of approximately <u>3.00×108 m/s</u>
The wavelength should have units of meters.
The frequency should have units of Hz or s−1. Hz is hertz or reciprocal seconds.
We know the value of the speed of light and we are given the wavelength. All we have to do is rearrange the equation to solve for the frequency:
<u>v = c/λ</u>
= 3.00× m/s/505×
m
= 5.94Xs-1
= <u>5.94X</u><u> Hz</u>
<u />
Frequency describes the number of waves passing through a particular location at a particular time. A class interval frequency is the number of observations that occur in a particular predefined interval. So, for example, if 20 of her ages 5-9 appear in the survey's data, the interval 5-9 has a frequency of 20. The class interval endpoints are the lowest and highest possible values of the variable.
Learn more about The frequency here:- brainly.com/question/254161
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The complete question is shown in the image attached to this answer.
Answer:
C
Explanation:
Let us quickly remember that the EMF of a cell under non standard conditions in given by the Nernst equation.
This equation states that;
E = E°cell - 0.592/n log Q
Where
E = EMF under non standard conditions
E°cell= standard EMF of the cell
n = number of electrons transferred
Q = reaction quotient
If the reaction quotient is greater than 1 then cell potential is less than the standard cell potential.
The cell that generates the lowest cell potential is the cell depicted in option C because Q has the greatest positive value(Q<1).
