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Aneli [31]
3 years ago
15

The Franck-Hertz experiment involved shooting electrons into a low-density gas of mercury atoms and observing discrete amounts o

f kinetic energy loss by the electrons. Suppose instead a similar experiment is done with a very cold gas of atomic hydrogen, so that all of the hydrogen atoms are initially in the ground state. If the kinetic energy of an electron is 11.1 eV just before it collides with a hydrogen atom, how much kinetic energy will the electron have just after it collides with and excites the hydrogen atom? Kfinal = eV
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Answer:

the final kinetic energy is 0.9eV

Explanation:

To find the kinetic energy of the electron just after the collision with hydrogen atoms you take into account that the energy of the electron in the hydrogen atoms are given by the expression:

E_n=\frac{-13.6eV}{n^2}

you can assume that the shot electron excites the electron of the hydrogen atom to the first excited state, that is

E_{n_2-n_1}=-13.6eV[\frac{1}{n_2^2}-\frac{1}{n_1^2}]\\\\E_{2-1}=-13.6eV[\frac{1}{2^2}-\frac{1}{1}]=-10.2eV

-10.2eV is the energy that the shot electron losses in the excitation of the electron of the hydrogen atom. Hence, the final kinetic energy of the shot electron after it has given -10.2eV of its energy is:

E_{k}=11.1eV-10.2eV=0.9eV

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Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find
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Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

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. On a safari, a team of naturalists sets out toward a research station located 9.6 km away in a direction 42° north of east. Af
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Answer:\theta =49.76^{\circ} North of east

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Given

Research station is 9.6 km away in 42^{\circ}North of east

after travelling 3.1 km 25^{\circ} north of east

Position vector of safari after 3.1 km is

r_2=3.1cos25\hat{i}+3.1sin25\hat{j}

Position vector if had traveled correctly is

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Now applying triangle law  of vector addition we can get the required vector(r_1)

r_1+r_2=r_0

r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}

r_1=4.325\hat{i}+5.112\hat{j}

Direction is given by

tan\theta =\frac{y}{x}=\frac{5.112}{4.325}

\theta =49.76^{\circ}

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