Question

An electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength becomes equal to 1.54 Å.

Answer 1

To determine: The potential required for a beam of electrons to be accelerated so that its wavelength becomes $1.54 \AA$.

Given Data:  $\lambda \quad =\quad 1.54\quad \times \quad { 10 }^{ -10 }$

Constants: Planck's constant, $h\quad =\quad 6.63\quad \times \quad { 10 }^{ -34 }\quad kg\sfrac { { m }^{ 2 } }{ s }$

Charge of an electron, $e\quad =\quad 1.602\quad \times \quad { 10 }^{ -19 }\quad C$

Mass of an electron $=\quad 9.1\quad \times \quad { 10 }^{ -31 }\quad kg$

Formulas to be used:

1) Kinetic energy of an electron in terms of mass m and speed 'u' of the electron $=\quad \frac { 1 }{ 2 } m{ u }^{ 2 }$

2) Kinetic energy of an electron in terms of potential, $V\quad =\quad e.V$

3) Wavelength, $\lambda \quad =\quad \frac { h }{ mu }$

Calculation:

Step 1: Equating the formulas (1) and (2), and substituting u in terms of     using formula (3) we get

$\frac { 1 }{ 2 } m{ u }^{ 2 } \quad =\quad e.V$

$\frac { 1 }{ 2 } m{ \left( \frac { h }{ m\lambda } \right) }^{ 2 }\quad =\quad e.V$

$\frac { 1 }{ 2 } \frac { { h }^{ 2 } }{ m{ \lambda }^{ 2 } } \quad =\quad e.V$

$V\quad =\quad \frac { 1\quad \times \quad \left( 6.63\quad \times \quad { 10 }^{ -34 } \right) }{ 2\quad \times \quad \left( 9.1\quad \times \quad { 10 }^{ -31 } \right) { \left( 1.54\quad \times \quad { 10 }^{ -10 } \right) }^{ 2 }\left( 1.602\quad \times \quad { 10 }^{ -19 } \right) } \quad =\quad 63.3\quad Volt$

Answer 2

Answer:

Hope this will help you!!