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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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Answer:

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Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

W_T+W_F=K_f-K_i (1)

Where:

W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to \frac{1}{2}mv^{2}, so if the initial velocity v_i is equal to zero, the initial kinetic energy K_i is equal to zero.

Then, replacing the values on the equation and solving for v_f, we get:

W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}

\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f

So, the speed after being pulled 3.2m is 2.4123 m/s

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