Question

What is the wavelength for the electron accelerated by $1.0 \times 10^{4}$ volts?

Answer 1

"Let's calculate the velocity of electron

$Energy\quad of\quad electron\quad =\quad 1.0\quad \times \quad { 10 }^{ 4 }\quad volts$

$E\quad =\quad 1.0\quad \times \quad { 10 }^{ 4 }\quad \times \quad 1.6\quad \times \quad { 10 }^{ -19 }\quad J$

$E\quad =\quad 1.6\quad \times \quad 1{ 0 }^{ -15 }\quad kg.{ m }^{ 2 }{ s }^{ -2 }$

Or

$\frac { 1 }{ 2 } m{ v }^{ 2 }\quad =\quad 1.6\quad \times \quad 1{ 0 }^{ -15 }\quad kg.{ m }^{ 2 }{ s }^{ -2 }$

Or

$v\quad =\quad \left( \frac { 2\quad \times \quad 1.6\quad \times \quad { 10 }^{ -15 }\quad kg.{ m }^{ 2 }{ s }^{ -2} }{ 9.1\quad \times \quad { 10 }^{ -31 }\quad kg } \right) \quad =\quad 5.93\quad \times \quad { 10 }^{ 7 }\quad { ms }^{ -1 }$

Let's calculate the wavelength of an electron

According to de Broglie equation,

$\lambda \quad =\quad \frac { h }{ mv }$

$\lambda \quad =\quad \frac { (6.626\quad \times \quad { 10 }^{ -34 }\quad kg.{ m }^{ 2 }{ s }^{ -1 }) }{ (9.1\quad \times \quad { 10 }^{ -31 }\quad kg)\quad \times \quad (5.93\quad \times \quad { 10 }^{ 7 }\quad m{ s }^{ -1 }) }$

$\lambda \quad =\quad 1.22\quad \times \quad { 10 }^{ -11 }\quad m$"

Answer 2

lamda = 1.22×10^(-11)

is the _____Answer