Answer:
1.907 x 10⁻⁵ J.
Explanation:
Given,
Volume of space, V = 5.20 m³
Assuming the intensity of sunlight(S) be equal to 1.1 x 10³ W/m².
Electromagnetic energy = ?
where c is the speed of light.
Hence, Electromagnetic energy is equal to 1.907 x 10⁻⁵ J.
Least count of the pulse stopwatch is given by
this means each division of the stopwatch will measure 0.1 s of time
After 3 journeys from one end to other we can see that total time that is measured here is shown by the clock as 52nd division
So here total time is given as
Time = (Number of division) (Least count)
now we will have
Answer:
(a) The energy of the photon is 1.632 x J.
(b) The wavelength of the photon is 1.2 x m.
(c) The frequency of the photon is 2.47 x Hz.
Explanation:
Let;
= -13.60 ev
= -3.40 ev
(a) Energy of the emitted photon can be determined as;
-
= -3.40 - (-13.60)
= -3.40 + 13.60
= 10.20 eV
= 10.20(1.6 x )
-
= 1.632 x
Joules
The energy of the emitted photon is 10.20 eV (or 1.632 x Joules).
(b) The wavelength, λ, can be determined as;
E = (hc)/ λ
where: E is the energy of the photon, h is the Planck's constant (6.6 x Js), c is the speed of light (3 x
m/s) and λ is the wavelength.
10.20(1.6 x ) = (6.6 x
* 3 x
)/ λ
λ =
= 1.213 x
Wavelength of the photon is 1.2 x m.
(c) The frequency can be determined by;
E = hf
where f is the frequency of the photon.
1.632 x = 6.6 x
x f
f =
= 2.47 x Hz
Frequency of the emitted photon is 2.47 x Hz.