1) n=1 -> n=2 : delta E = -5+11 = 6. The answer is D
<span>2) n=1 -> n=3 : delta E = -2+11 = 9. The answer is B </span>
<span>3) n=1 -> n=4 : delta E = -1+11 = 10. No solution available </span>
<span>4) n=1 -> infinity delta E = 11. The answer is A </span>
<span>5) not absorbed would be C, as there is no transition with delta E of 8. </span>
The speed of an object in a set direction is its velocity.
Answer:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is

Re-arranging for r

And substituting into the previous equation

Solving for T,

So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
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