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 recoiling 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.

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

 ---------  ( 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

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft