Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

Question

Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be: (A)  0.55 overset texto mathop textA  (B)  0.055 overset texto mathop textA  (C)  0.077 overset texto mathop textA  (D)  0.095 overset texto mathop textA

Tardigrade Admin · Accepted Answer

de Broglie wavelength λ =(12.27/√V) overset texto mathop textA λ =(12.27/√50000)=0.055 overset texto mathop textA

Q. Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

$0.55 A o$

$0.055 A o$

$0.077 A o$

$0.095 A o$

de Broglie wavelength $λ = \frac{12.27}{V} A o$ $λ = \frac{12.27}{50000} = 0.055 A o$

Q. Acceleration voltage in an electron gun is 50000 volt, de-Brogile wavelength of electron will be:

0.55Ao

0.055Ao

0.077Ao

0.095Ao

de Broglie wavelength λ=V​12.27​Ao λ=50000​12.27​=0.055Ao

$0.55 A o$

$0.055 A o$

$0.077 A o$

$0.095 A o$

de Broglie wavelength $λ = \frac{12.27}{V} A o$ $λ = \frac{12.27}{50000} = 0.055 A o$