De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(| e | = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g, h = 6.6 \times 10^{- 34} J s) .$
$0.5 ˚ A$
$1.7 ˚ A$
$2.4 ˚ A$
$1.2 ˚ A$

Question

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(| e | = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g, h = 6.6 \times 10^{- 34} J s) .$
$0.5 ˚ A$
$1.7 ˚ A$
$2.4 ˚ A$
$1.2 ˚ A$

Toppr Admin · Accepted Answer

-x3BB-hp-h-x221A-2mE-h-x221A-2mqV-x3BB-6-6-xD7-10-x2212-34-x221A-2-xD7-9-1-xD7-10-x2212-31-xD7-1-6-xD7-10-x2212-19-xD7-50-1-72-xD7-10-x2212-10-1-72A-xA0-Option B

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: (|e|=1.6×10−19C,me=9.1×10−31kg,h=6.6×10−34Js). 0.5˚A1.7˚A2.4˚A1.2˚A

λ=hp=h√2mE=h√2mqVλ=6.6×10−34√2×9.1×10−31×1.6×10−19×50=1.72×10−10=1.72AOption B

De-Broglie wavelength of an electron accelerated by a voltage of 50 V is close to: $(| e | = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g, h = 6.6 \times 10^{- 34} J s) .$
$0.5 ˚ A$
$1.7 ˚ A$
$2.4 ˚ A$
$1.2 ˚ A$

$λ = \frac{h}{p} = \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 m q V}}$
$λ = \frac{6.6 \times 10^{- 34}}{\sqrt{2 \times 9.1 \times 10^{- 31} \times 1.6 \times 10^{- 19} \times 50}} = 1.72 \times 10^{- 10} = 1.72 A$
Option B