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Usimov [2.4K]
3 years ago
15

In 1995 a research group led by Eric Cornell and Carl Wiemann at the University of Colorado successfully cooled Rubidium atoms t

o the 20-200 nK temperature range. Assuming (incorrectly) that the Rubidium atoms behave like particles of a classical ideal gas, calculate the RMS speed of a Rubidium atom at a temperature of 85.0 nK. In the experiments one particular isotope of Rubidium was used, Rubidium-87. The molar mass of this isotope is 86.91 g/mol.
Physics
1 answer:
saveliy_v [14]3 years ago
6 0

Answer:

0.00493 m/s

Explanation:

T = Temperature of the isotope = 85 nK

R = Gas constant = 8.341 J/mol K

M = Molar mass of isotope = 86.91 g/mol

Root Mean Square speed is given by

v_r=\sqrt{\dfrac{3RT}{M}}\\\Rightarrow v_r=\sqrt{\dfrac{3\times 8.314\times 85\times 10^{-9}}{86.91\times 10^{-3}}}\\\Rightarrow v_r=0.00493\ m/s

The Root Mean Square speed is 0.00493 m/s

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A proton is projected in the positive x direction into a region of a uniform electric field E S 5 126.00 3 105 2 i^ N/C at t 5 0
Dafna11 [192]

Answer:

(a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

Explanation:

Given that,

Electric field E=-6.00\times10^{5}i\ N/C

Time = 5.0 sec

Distance 7.00 cm

(a). We need to calculate the acceleration

Using formula of force

F=F_{e}

ma=qE

a=\dfrac{Eq}{m}

Where, E = electric field

m = mass of proton

q = charge of proton

Put the value into the formula

a=\dfrac{-6.00\times10^{5}\times1.6\times10^{-19}}{1.67\times10^{-27}}

a=-5.74\times10^{13}\ m/s62

The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

Using equation of motion

v^2-u^2=2as

Where, s = distance

Put the value into the formula

0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

u=\sqrt{2\times5.74\times10^{13}\times7.00\times10^{-2}}

u=2834783.94

u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

Using formula

t=\dfrac{u}{a}

Where, u = initial velocity

a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

5 0
3 years ago
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