Question

Calculate the wavelength associated with a moving electron having kinetic energy of 1.375×10 to power minus 25 J . plz solve this systematic way​

Answer 1

Answer :-

Wavelength associated with the moving electron is 1.32 × 10⁻⁶ metres .

Explanation :-

We have :-

→ Kinectic energy (K.E.) = 1.375 × 10⁻²⁵ J

→ Planck's Constant (h) = 6.62 × 10⁻³⁴ Js

→ Mass of electron (m) = 9.1 × 10⁻³¹ kg

To find :-

→ Wavelength (λ) .

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Firstly, let's calculate the velocity of the electron .

K.E. = ½mv²

⇒ 1.375 × 10⁻²⁵ = ½ × 9.1 × 10⁻³¹ × v²

⇒ v² = (1.375 × 10⁻²⁵ × 2)/(9.1 × 10⁻³¹)

⇒ v² = (2.75 × 10⁻²⁵)/(9.1 × 10⁻³¹)

⇒ v² = 2.75/9.1 × 10³¹⁻²⁵

⇒ v² = 0.302 × 10⁶

⇒ v = √302000

⇒ v = 549.54 m/s

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Now, wavelength associated will be :-

λ = h/mv

⇒ λ = (6.62 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 549.54)

⇒ λ = (6.62/5000.81) × 10³¹⁻³⁴

⇒ λ = 0.00132 × 10⁻³

⇒ λ = 1.32 × 10⁻⁶ m

Answer 2

Given :-

The Kinetic energy of 1.375 × 10⁻²⁵ J

To Find :-

Calculate the wavelength

Solution :-

We know that

$\sf \lambda = \dfrac{h}{\sqrt{2\times m\times KE}}$

m = 9.31 × 10⁻³¹

$\sf :\implies \dfrac{6.62\times10^{-34}}{\sqrt{2\times9.31\times 10^{-31}\times 1.375\times 10^{-25}}}$

$\sf :\implies \dfrac{6.62\times 10^{-34}}{\sqrt{25.6025\times 10^{-31-25}}}$

$\sf :\implies \dfrac{6.62\times 10^{-34}}{\sqrt{25.6025\times 10^{-56}}}$

$\sf :\implies \dfrac{6.62\times 10^{-34}}{5.05\times10^{-40}}$

$\sf :\implies \dfrac{6.62\times10^{-34}\times10^{40}}{5.05}$

$\sf :\implies 1.31 \times 10^{-6}\; m$