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bagirrra123 [75]
3 years ago
10

What is frequency of light is emitted when an electron jumps into the smallest orbit of hydrogen, coming from a very large radiu

s (assume infinity)?
Physics
1 answer:
mrs_skeptik [129]3 years ago
7 0

Answer:

3.28403\times 10^{15}\ Hz

Explanation:

n_1 = 1

n_2=\infty

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

The energy is given by

E(n)=E_1(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\\\Rightarrow E(1)=13.6\ eV

Energy is also given by

E=hf\\\Rightarrow f=\dfrac{E}{h}\\\Rightarrow f=\dfrac{13.6\times 1.6\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=3.28403\times 10^{15}\ Hz

The frequency of light emitted would be 3.28403\times 10^{15}\ Hz

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Answer:

The horizontal distance is 0.64 m.

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Answer:

Part a)

t = 2.83 s

Part b)

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Part a)

Since both the balls are projected with same speed in opposite directions

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Afterwards the motion will be same as the first ball which is projected downwards

so here the time difference is given as

\Delta y = 0 = v_y t + \frac{1}{2}at^2

0 = 13.9 t - \frac{1}{2}(9.81) t^2

t = 2.83 s

Part b)

Since the displacement in y direction for two balls is same as well as the the initial speed is also same so final speed is also same for both the balls

so it is given as

v_f^2 - v_i^2 = 2 a \Delta y

v_f^2 - (13.9)^2 = (2)(-9.81)(-19.1)

v_f^2 = 567.9

v_f = 23.8 m/s

Part c)

Relative speed of two balls is given as

v_{12} = v_1 - v_2

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now the distance between two balls in 0.8 s is given as

d = v_{12} t

d = 27.8 \times 0.8

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