The de-Broglie wavelength of a proton (charge =1.6× 10-19C .mass =1.6× 10-27kg ) accelerated through a potential difference of 1 kV is

Question

The de-Broglie wavelength of a proton (charge =1.6× 10-19C .mass =1.6× 10-27kg ) accelerated through a potential difference of 1 kV is (A)  600 mathringA (B)  0.9× 10-12m  (C)  7 mathringA (D) 0.9 nm

Tardigrade Admin · Accepted Answer

According to de-Broglie hypothesis λ =(h/p) =(h/√2mE)=(h/√2mqV) ∴ λ =(6.6× 10-34/√2× (1.6× 10-27)(1.6× 10-19)× 1000) =(6.6× 10-34/7.16× 10-22) =0.9× 10-12m

Q. The de-Broglie wavelength of a proton (charge $= 1.6 \times 10^{- 19} C$ .mass $= 1.6 \times 10^{- 27} k g$ ) accelerated through a potential difference of $1 kV$ is

$600 \overset{˚}{A}$

$0.9 \times 10^{- 12} m$

$7 \overset{˚}{A}$

0.9 nm

According to de-Broglie hypothesis
$λ = \frac{h}{p}$
$= \frac{h}{2 m E} = \frac{h}{2 m q V}$
$∴$ $λ = \frac{6.6 \times 10 ^{- 34}}{2 \times ( 1.6 \times 10 ^{- 27} ) ( 1.6 \times 10 ^{- 19} ) \times 1000}$
$= \frac{6.6 \times 10 ^{- 34}}{7.16 \times 10 ^{- 22}}$
$= 0.9 \times 10^{- 12} m$

Q. The de-Broglie wavelength of a proton (charge =1.6×10−19C .mass =1.6×10−27kg ) accelerated through a potential difference of 1kV is

600A˚

0.9×10−12m

7A˚