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exis [7]
3 years ago
5

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

ed state, emitting a 466 A photon. Find a) The energy of the free electron, and b) The energy of the original photon
Physics
1 answer:
anygoal [31]3 years ago
3 0

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}

            = 26.6 e V

h c = 6.626 ×  10⁻³⁴ ×  3  × 10⁸  = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁  = 23.2 + 13.6

       = 36.8 e V

energy of the original photon is 36.8 eV

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q = 1.22y/D

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5 0
3 years ago
A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total
Lynna [10]

Explanation:

Given Data

Total mass=93.5 kg

Rock mass=0.310 kg

Initially wagon speed=0.540 m/s

rock speed=16.5 m/s

To Find

The speed of the wagon

Solution

As the wagon rolls, momentum is given as

P=mv

where

m is mass

v is speed

put the values

P=93.5kg × 0.540 m/s

P =50.49 kg×m/s

Now we have to find the momentum of rock

momentum of rock = mv

momentum of rock = (0.310kg)×(16.5 m/s)

momentum of rock =5.115 kg×m/s  

From the conservation of momentum we can find the wagons momentum So

wagon momentum=50.49 -5.115 = 45.375 kg×m/s  

Speed of wagon = wagon momentum/(total mass-rock mass)

Speed of wagon=45.375/(93.5-0.310)

Speed of wagon= 0.487 m/s

Throwing rock backward,

momentum of wagon = 50.49+5.115 = 55.605  kg×m/s

Speed of wagon = wagon momentum/(total mass-rock mass)

speed of wagon = 55.605  kg×m/s/(93.5kg-0.310kg)

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7 0
2 years ago
a spring with a constant of 80N/m is stretched by a force of 240N. how much the displacement of the spring from equilibrium?
Inessa05 [86]

Answer:

1200N/m

Explanation:

given parameters:

force on the motorcycle spring is 240N

Extension 2cm or 0.02m

unknown _

spring constant:

:?

solution:

to a spring a force applied is given as :

f=ke

f is applied as force

k is spring constant

e is the Extension

240= kx0.02

k=1200N/m

8 0
2 years ago
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