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Alja [10]
2 years ago
5

When the potential difference between the plates of a capacitor is increased by 3.50 V , the magnitude of the charge on each pla

te increases by 15.0 μC . What is the capacitance of this capacitor in μF?
Physics
1 answer:
hodyreva [135]2 years ago
5 0

Answer:

42.9 μF

Explanation:

V = 3.50 V, Q = 150 μC

C = Q/V = 150/3.50 μF = 42.9 μF

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A. 2.82 eV

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C. 59.5 angstroms

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n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

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c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

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

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r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

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