Answer:
the light emitting must be of greater wavelength
Explanation:
For this exercise we must use the Planck equation
E = h f
And the speed of light
c = λ f
f = c / λ
We replace
E = h c / λ
The wavelength of the green light is of the order of 500 nm, let's calculate the energy
E = 6.63 10⁻³⁴ 3 10⁸ /λ
E = 1,989 10⁻²⁵ /λ
λ = 500 nm = 500 10⁻⁹ m
E = 1,989 10⁻²⁵ / 500 10⁻⁹
E = 3,978 10⁻¹⁹ J
That is the energy of the transition for a transition is an intermediate state the energy must be less, this implies that the wavelength must increase. For the explicit case of a state with half of this energy
= E / 2
= 3,978 10⁻¹⁹ / 2 = 1,989 10⁻¹⁹
Let's clear and calculate
λ = h c / E
λ = 1,989 10⁻²⁵ / 1,989 10⁻¹⁹
λ = 1 10⁻⁶ m
Let's reduce to nm
λ = 1000 nm
This wavelength is in the infrared region
the light emitting must be of greater wavelength
Answer:
The kinetic energy is: 50[J]
Explanation:
The ball is having a potential energy of 100 [J], therefore
PE = [J]
The elevation is 10 [m], and at this point the ball is having only potential energy, the kinetic energy is zero.
![E_{p} =m*g*h\\where:\\g= gravity[m/s^{2} ]\\m = mass [kg]\\m= \frac{E_{p} }{g*h}\\ m= \frac{100}{9.81*10}\\\\m= 1.01[kg]\\\\](https://tex.z-dn.net/?f=E_%7Bp%7D%20%3Dm%2Ag%2Ah%5C%5Cwhere%3A%5C%5Cg%3D%20gravity%5Bm%2Fs%5E%7B2%7D%20%5D%5C%5Cm%20%3D%20mass%20%5Bkg%5D%5C%5Cm%3D%20%5Cfrac%7BE_%7Bp%7D%20%7D%7Bg%2Ah%7D%5C%5C%20m%3D%20%5Cfrac%7B100%7D%7B9.81%2A10%7D%5C%5C%5C%5Cm%3D%201.01%5Bkg%5D%5C%5C%5C%5C)
In the moment when the ball starts to fall, it will lose potential energy and the potential energy will be transforme in kinetic energy.
When the elevation is 5 [m], we have a potential energy of
![P_{e} =m*g*h\\P_{e} =1.01*9.81*5\\\\P_{e} = 50 [J]\\](https://tex.z-dn.net/?f=P_%7Be%7D%20%3Dm%2Ag%2Ah%5C%5CP_%7Be%7D%20%3D1.01%2A9.81%2A5%5C%5C%5C%5CP_%7Be%7D%20%3D%2050%20%5BJ%5D%5C%5C)
This energy is equal to the kinetic energy, therefore
Ke= 50 [J]
Answer:
Person B has four times the power output of person A.
In transistor,
Emitter current is equal to the sum of base current and collector current.
Thanks!
Answer:
option E
Explanation:
given,
diameter = 4 mm
shutter speed = 1/1000 s
diameter of aperture = ?
shutter speed = 1/250 s
exposure time to the shutter time
N is the diameter of the aperture and t is the time of exposure
now,
inserting all the values
N₂² = 4
N₂ = 2 mm
hence , the correct answer is option E
