A beam of electrons ( m 0 = 9.11 × 1 0 − 31 kg) moving at 2 × 1 0 6 m/s strikes an area of 1 m m 2 at the rate of 2 × 1 0 14 per second and is absorbed. Find the pressure in units of 1 0 − 4 Pa (two decimals).

Pressure = A N m 0 v = 1 0 − 6 2 × 1 0 14 × 9.11 × 1 0 − 31 × 2 × 1 0 6 = 3.64 × 1 0 − 4 Pa.

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Physics - Heat & Thermo Question with Solution | TestHub

PhysicsHeat & ThermoKinetic Theory of Gases and Gas LawsMedium2 minQB

PhysicsMediumnumerical

A beam of electrons ( $m_{0} = 9.11 \times 1 0^{- 31}$ kg) moving at $2 \times 1 0^{6}$ m/s strikes an area of 1 $m m^{2}$ at the rate of $2 \times 1 0^{14}$ per second and is absorbed. Find the pressure in units of $1 0^{- 4}$ Pa (two decimals).

Answer:

3.64

Solution:

Pressure =

$\frac{N m _{0} v}{A} = \frac{2 \times 1 0 ^{14} \times 9.11 \times 1 0 ^{- 31} \times 2 \times 1 0 ^{6}}{1 0 ^{- 6}} = 3.64 \times 1 0^{- 4}$

Pa.

Subject:PhysicsTopic:Heat & ThermoSubtopic:Kinetic Theory of Gases and Gas Laws

⏱ 2mℹ️ Source: QB

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