Answer:
A. 2.82 eV
B. 439nm
C. 59.5 angstroms
Explanation:
A. To calculate the energy of the photon emitted you use the following formula:
(1)
n1: final state = 5
n2: initial state = 2
Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

B. The energy of the emitted photon is given by the following formula:
(2)
h: Planck's constant = 6.62*10^{-34} kgm^2/s
c: speed of light = 3*10^8 m/s
λ: wavelength of the photon
You first convert the energy from eV to J:

Next, you use the equation (2) and solve for λ:

C. The radius of the orbit is given by:
(3)
where ao is the Bohr's radius = 2.380 Angstroms
You use the equation (3) with n=5:

hence, the radius of the atom in its 5-th state is 59.5 anstrongs
Its C because if it is a low frequency it will not change much so it will be a longer wavelength and the higher the frequency the shorter the wavelength
If object is not accelerating, the sum of all forces on the object will be equal to ZERO...
Answer:
D
Explanation:
She says that the object of the experiment is to see how far the string stretches given a mass attached to the string.
The only thing that is at issue is either the mass or the amount the string stretches.
Nothing else matters.
The dependent variable therefore is the amount the string stretches. So the last choice is the answer.