The equation λ=(1.227/ x ) nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for: Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V= Accelerating potential in volts for electron

Question

The equation λ=(1.227/ x ) nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for: Where, m = mass of electron P = momentum of electron K = Kinetic energy of electron V= Accelerating potential in volts for electron (A) √m K (B) √ P  (C) √ K  (D) √ V

Tardigrade Admin · Accepted Answer

.λ=( h / m v) text (de-Broglie's wavelength ) λ ( h /√2 m ( K ⋅ E )) h =( h /√2 mqV ) Putting the values of m ; q We get λ=(1.22/√ V ) nm

Q. The equation $λ = \frac{1.227}{x} nm$ can be used to find the de-Brogli wavelength of an electron.
In this equation $x$ stands for :
Where,
$m =$ mass of electron
$P =$ momentum of electron
$K =$ Kinetic energy of electron
$V =$ Accelerating potential in volts for electron

$m K$

$P$

$K$

$V$

$λ = \frac{h}{m v} (de-Broglie’s wavelength)$
$λ \frac{h}{2 m ( K \cdot E )}$
$h = \frac{h}{2 m q V}$
Putting the values of $m; q$
We get $λ = \frac{1.22}{V} nm$

Q. The equation λ=x1.227​nm can be used to find the de-Brogli wavelength of an electron. In this equation x stands for : Where, m= mass of electron P= momentum of electron K= Kinetic energy of electron V= Accelerating potential in volts for electron

mK​

P​

K​

V​

λ=mvh​ (de-Broglie’s wavelength ) λ2m(K⋅E)​h​ h=2mqV​h​ Putting the values of m;q We get λ=V​1.22​nm

Q. The equation $λ = \frac{1.227}{x} nm$ can be used to find the de-Brogli wavelength of an electron.
In this equation $x$ stands for :
Where,
$m =$ mass of electron
$P =$ momentum of electron
$K =$ Kinetic energy of electron
$V =$ Accelerating potential in volts for electron

$m K$

$P$

$K$

$V$

$λ = \frac{h}{m v} (de-Broglie’s wavelength)$
$λ \frac{h}{2 m ( K \cdot E )}$
$h = \frac{h}{2 m q V}$
Putting the values of $m; q$
We get $λ = \frac{1.22}{V} nm$