A 20.0-kg mass is traveling to the right with a speed of 3.00 m/s on a smooth horizontal surface when it collides with and stick
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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