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Black_prince [1.1K]
3 years ago
10

When the tube is filled with mercury vapor, as in this case, a sharp drop in the collected current is observed when the accelera

ting potential reaches 4.9 VV. The current drops suddenly to a negligible value and slowly increases again when the accelerating potential is above 4.9 VV. What is the energy EEE absorbed by the atomic electrons in the mercury atom when the accelerating potential is 4.9 VV
Physics
1 answer:
adell [148]3 years ago
3 0

Answer:

4.9 eV .

Explanation:

It is the case of discharge through mercury tube light . In it , mercury  atoms are exited due to which electrons are sent to higher energy  level . Here current drops to zero because electrons are excited to higher level . Energy are absorbed in quantised manner . Energy absorbed by electrons will be 4.9 eV . That means , difference in energy  between two energy level is  4.9 eV .

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When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart
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Explanation:

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Radius of the ring (r) = 10.0 cm = 0.10 m           [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = 75\times 10^{-6}\ \mu C    [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}

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(a)

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C

(b)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C

(c)

Given:

Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2

Electric field is given as:

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Electric field is given as:

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Answer:

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