Answer:
E = 1.14 x 10⁻¹⁵ J
T for H = 53 K = 326 ºC
T for Xe = 0.40 K
Step-by-step 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.