The question is incomplete, here is the complete question:
Which is greater, the energy of one photon of orange light or the energy of one quantum of radiation having a wavelength of [tex]3.36\times 10^{-9}m[/tex]
Answer: The energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
Explanation:
To calculate the energy of one photon, we use the Planck's equation:
[tex]E=\frac{N_Ahc}{\lambda}[/tex]
where,
E = energy of radiation
[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}mol^{-1}[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
c = speed of light = [tex]3\times 10^8 m/s[/tex]
[tex]\lambda}[/tex] = wavelength of radiation
For orange light:For 1 photon, the term [tex]N_A[/tex] does not appear
[tex]\lambda}[/tex] = 620 nm = [tex]620\times 10^{-9}m[/tex] (Conversion factor: [tex]1nm=10^{-9}m[/tex] )
Putting values in above equation, we get:
[tex]E=\frac{6.626\times 10^{-34}Js\times 3\times 10^8m/s}{620\times 10^{-9}m}\\\\E=3.206\times 10^{-19}J[/tex]
For one quantum of radiation:[tex]\lambda}[/tex] = [tex]3.36\times 10^{-9}m[/tex]
Putting values in above equation, we get:
[tex]E=\frac{6.022\times 10^{23}mol^{-1}\times 6.626\times 10^{-34}Js\times 3\times 10^8m/s}{3.36\times 10^{-9}m}\\\\E=3.56\times 10^{7}J/mol[/tex]
Hence, the energy of one quantum of radiation having wavelength [tex]3.36\times 10^{-9}m[/tex] is greater than the energy of 1 photon of orange light.
A 22.0 kg child is riding a playground merry-go- round that is rotating at 40.0 rev/min. What centripetal force must
Answer:
F = 482.51 N
Explanation:
Given that,
Mass of a child, m = 22 kg
Angular velocity of the merry-go-round, [tex]\omega=40\ rev/min[/tex]
Let the radius of the path, r = 1.25 m
We need to find the centripetal force acting on the child. The formula for the centripetal force is given by :
[tex]F=m\omega^2r\\\\=22\times (4.18879)^2\times 1.25\\\\=482.51\ N[/tex]
So, the required centripetal force is 482.51 N.
An electron traverses a vacuum tube with a length of 2 m in 2 X 10- 4
sec. What is the average speed of the
electron during this time?
Answer:
Average speed = 10,000 m/s
Explanation:
Given the following data;
Distance = 2m
Time = 0.0002secs
To find the average speed;
Average speed = distance/time
Average speed = 2/0.0002
Average speed = 10,000 m/s
Therefore, the average speed of the
electron is 10,000 meters per seconds.
The human nervous system can propagate nerve impulses at about 102 m>s. Estimate the time it takes for a nerve impulse to travel 2 m from your toes to your brain.
Answer:
t = 0.196 s
Explanation:
The speed of a pulse is determined by the characteristics of the medium, its density and its resistance to stress, as long as these remain the speed will be constant for which we can use the kinetic expressions of the uniform movement
v = x / t
t = x / v
calculate
t = 2/102
t = 0.196 s
Allen and Jason are chucking a speaker around. On one particular throw, Allen throws the speaker, which is playing a pure tone of frequency f, at a speed of 10 m/s directly towards Jason, but his aim is a bit off. As a result, Jason runs forward towards the speaker at a speed of 6 m/s before catching it. Then, the frequency that Jason hears while running can be written as (m/n)f Hz, where m and n are relatively prime positive integers. Compute m n.
Answer:
Explanation:
We shall apply Doppler's effect of sound .
speaker is the source , Jason is the observer . Source is moving at 10 m /s , observer is moving at 6 m/s .
apparent frequency = [tex]f_o\times\frac{V+v_o}{ V-v_s}[/tex]
V is velocity of sound , v₀ is velocity of observer and v_s is velocity of source and f_o is real frequency of source .
Here V = 340 m/s , v₀ is 6 m/s , v_s is 10 m/s . f_o = f
apparent frequency = [tex]f\times \frac{340+6}{340-10}[/tex]
= [tex]f\times \frac{346}{330}[/tex]
So m = 346 , n = 330 .