Answer:
Speed of electron = 4.15 X 10raised to power 6 m/s
Gas temperature of hydrogen = 26K
Gas temperature of Xenon = 0.39K
Explanation:
Louis De brogile was a french physicist who was the first to make the postulation on wave been a beam of light, it says that waves transfers energy and momentum via photons. he came about the equation which shows the relationship between wavelength (λ), momentum(p) and the plancks's constant (h). Mathematically from de brogile equation;
λ = h/p , where λ is wavelength metre, p is momentum in kgm/s and h is planck's constant in J/s.
but momentum is mass X velocity, hence λ = h/mv, hence the popular de-brogile equation.
The root mean square speed shows the connection between the average kinetic energy of molecules and the gas temperature. it is derived from the ideal gas equation and the average kinetic energy. Vrms = square root ( 3RT/M).
Xe atomic beams are not suitable for atomic diffraction experiments because of its low gas temperature as it is seen from the calculation, and also because of its high relative atomic mass.
Answer:
E = 1.14 x 10⁻¹⁵ J
T for H = 53 K = 326 ºC
T for Xe = 0.40 K
Explanation:
The strategy here is to use the formula E = h c/λ for the first question.
h = plancks constant, 6.626 x 10⁻³⁴ Js
c = speed of light, 3 x 10⁸ m/s
λ = 0.175 nm = 0.175 nm x ( 1 m/ 10⁹ nm )= 1.75 x 10⁻¹⁰ m
Plugging the values:
E = 6.626 x 10⁻³⁴ Js x 3 x 10⁸ m/s / 1.75 x 10⁻¹⁰ m = 1.14 x 10⁻¹⁵ J
For the second part, we know that momentum p = h/ λ from De Broglies equation, and momentum is also mass times velocity.
This velocity will be then used to calculate the gas temperature from the relation vrms = √3kT/m for H, Xe
For the typical value of 0.35 nm needed to resolve the diffraction, we have:
p= h / λ = 6.626 x 10⁻³⁴ Js/ /( 3.5 x 10⁻¹⁰ m) = [1.9 x 10⁻²⁴( Kg m/s² x m) s]/ m
= 1.9 x 10⁻²⁴ Kg m/s ( Joules has units of force x distance)
Now p = mv ⇒ v = p/m
The mass in this expression is the atomic mass of H and Xe
m H = 1 g/ 6.022 ²³ atoms = 1.66 x 10⁻²⁴ g x 1 Kg/1000 g = 1.66 x 10⁻²⁷ Kg
m Xe = 131.30 g/ 6.022 x 10²³ = 2.18 x 10⁻²² g x 1Kg/1000g = 2.18 x 10⁻²⁵ kg
v H = 1.9 x 10⁻²⁴ Kg m/s / 1.66 x 10⁻²⁷ Kg = 1144.6 m/s
v Xe = 1.9 x 10⁻²⁴ Kg m/s / 2.18 x 10⁻²⁵ kg = 8.7 m/s
and to solve for the temperature at which λ is 0.35 nm:
vrms = √( 3 kT / m ) ⇒ T = m vrms² / 3k where k = Boltzman constant,
1.38 x 10⁻²³ J/K
T for H = 1.66 x 10⁻²⁷ Kg x ( 1144.6 m/s )² / (3 x 1.38 x 10⁻²³ J/K)
= 53 K = ( 53 +273 K ) = 326 ºC
T for Xe = 2.18 x 10⁻²⁵ kg x ( 8.7 m/s )² / (3 x 1.38 x 10⁻²³ J/K)
= 0.40 K !!!!!!
Therefore Xe is not suitable at all to work the diffraction at such low temperature, almost zero. To make matters even worse for Xe its freezing point is -111.9 º C = ( -111.9 + 273 ) K = 161.1 K, so we would not even can has the monoatomic gas beam for Xe.
