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JulsSmile [24]
3 years ago
15

A. If an electron of 16 eV had a head-on collision with a Cs atom at rest, what would be the kinetic energy in eV) of the recoil

ing Cs atom? Assume an elastic collision. (Take the atom to be the most abundant isotope of the element. B. In a Frank-Hertz experiment, the first excited state of Hg is obtained at an accelerating voltage measured to be 4.87 V. Based on this measured value, determine the wavelength of the ultraviolet radiation you expect to be emitted.
Physics
1 answer:
Fiesta28 [93]3 years ago
8 0

Answer:

A) 1.96 * 10^-8 J

B) 2.55 * 10^-7 m

Explanation:

A) calculate the kinetic energy in eV of the recoiling Cs atom

we have to apply the principle of energy conservation and momentum

Initial kinetic energy of electron = 16 eV = 16 * (1.6 * 10^-19 ) J

Initial kinetic energy of atom = 0

therefore the final kinetic energy after collision ( E )

= [ 16 * ( 1.6 * 10^-19 ) ] + 0

= 1.96 * 10^-8 J

B) Determine the wavelength of the ultraviolet radiation

accelerating Voltage = 4.87 V

K.E = eV = 1.6 * 10^-19 * 4.87

next we will apply Planck's relationship

\frac{hc}{w} = KE  ---------  ( 1 )

w = wavelength

h = 6.64 * 10^-34 J-S ( Planck's constant )

c = 3 * 10 ^8 m/sec  ( speed of light )

KE = 1.6 * 10^-19 * 4.87

substitute given values into equation 1 above

w ( wavelength ) =  2.55 * 10^-7 m

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A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
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Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

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