91Ó°ÊÓ

• Angle $\mathrm{θ}=55°$
• Speed $v=1.0km/s$
• Mass $m=3.3×{10}^{-24}g$
• Area $A=2.0c{m}^2$
• Rate of strike is ${10}^{23}\text{molecules}/\text{sec}$

The expression for impulse is given by,

$J=F×t$

Here J is the impulse, F is the force and T is the time.

The expression for pressure is given by,

$P=\frac{F}{A}$

Here is the pressure, F is the force and A is the area.

The change in momentum by hydrogen moleculenormal tothewall is as follows:

$\mathrm{Δ}p=2Nmv cos\mathrm{θ}$ …… (i)

Here is the mass of hydrogen molecule, v is the speed of hydrogen molecule, and n is Avogadro’s number.

Now, we have to use the formula for impulse.

$\text{J}=\text{F}×\text{t}=\mathrm{Δ}\text{p}$…… (ii)

From equation (i) and (ii)

$F×t=2Nmv \text{cos}\mathrm{θ}$

Thus,

$F=\frac{2Nmv cos\mathrm{θ}}{t}$

Here, N/t is the rate at which the beam of the hydrogen molecules strike the wall.

So,

$$\begin{gathered} F=2×{10}^{23}×3.3×{10}^{-27}×{10}^3×cos55° \\ ⇒F=0.3785N \end{gathered}$$

Now, pressure is given as,

$P=\frac{F}{A}$

Thus,

$P=\frac{0.3785}{2×{10}^{-4}}$

On simplifying the above expression,

$$\begin{gathered} P=1892.8Pa \\ =1.89kPa \\ ≈1.9kPa \\ =1.9×{10}^{3}Pa \end{gathered}$$

Therefore, the beam’s pressure on the wall is$1.9×{10}^{3}Pa$ .