Question

the kinetic energy of an electron is 4.55 x 10-25 j. the mass of electron is 9.1 x 10-31 kg. calculate velocity of the electron

Answer 1

therefore, the velocity of the electron =1000m/s

Answer 2

$\huge {\sf{ {\underline {\underline{Given : - }}}}}$

• Kinetic energy of electron is 4.55 × 10⁻²⁵ J

• Mass of electron is 9.1 × 10⁻³¹ kg

$\huge {\sf { {\underline {\underline{To \: find : - }}}} }$

• Velocity of the electron

$\huge {\sf{ {\underline {\underline{Solution : - }}}}}$

Kinetic energy for a body is give by,

$\large {\underline {\bold {\boxed {\bigstar{ \: KE = \dfrac{1}{2} m {v}^{2} }}}}}$

Where ,

• m is mass of the body

• v is velocity of the body

We have ,

• m = 9.1 × 10⁻³¹ kg

• KE = 4.55 × 10⁻²⁵ J

$\sf: \implies4.55 \times {10}^{ - 25} = \frac{1}{2} \times 9.1 \times {10}^{ - 31} \times {v}^{2} \\ \\ \sf : \implies2 \times 4.55 \times {10}^{ - 25} = 9.1 \times {10}^{ - 31} \times {v}^{2} \\ \\ \sf: \implies9.1 \times 10 {}^{ - 25} = 9.1 \times {10}^{ - 31} \times {v}^{2} \\ \\ \sf: \implies \: {v}^{2} = \dfrac{9.1 \times {10}^{ - 25} }{9.1 \times 10 {}^{ - 31} } \\ \\ \sf: \implies \: {v}^{2} = \dfrac{ \cancel{9.1} \times {10}^{ - 25} }{ \cancel{9.1} \times {10}^{ - 34} } \\ \\ \sf : \implies \: {v}^{2} = \dfrac{ {10}^{ - 25} }{ {10}^{ - 31} } \\ \\ \sf : \implies \: {v}^{2} = {10}^{6} \\ \\ \sf : \implies \: v = \sqrt{10 {}^{6} } \\ \\ \sf : \implies \: v = \sqrt{ {(10)}^{3 {}^{(2)} } } \\ \\ \sf : \implies v = {10}^{3} \\ \\ \sf: \implies \: v = 1000 \: m {s}^{ - 1}$

∴ The velocity of the electron is 1000 m/s

$\huge{\sf{\underline{\underline{Additional\:Info:-}}}}$

❃ Potential energy of a body is given by ,

$\large\sf{\underline{\bold{\boxed{\bigstar{PE=mgh}}}}}$

Where ,

• m is mass of the body

• v is velocity of the body

• g is acceleration due to gravity

❃ Kinetic energy is always positive and is independent of direction (Scalar quantity)

❃ Kinetic energy depends on the frame of reference.