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Answer :
Answer:
The value is [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]
Explanation:
From the question we are told that
The velocity of the electron is [tex]v = 1.37 *10^{6} \ m/s[/tex]
The mass of the electron is [tex]m = 9.11 *10^{-28} \ g = 9.11 *10^{-31} \ kg[/tex]
Generally the deBroglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{mv}[/tex]
Here h is the Planck'c constant with value [tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34} }{9.11 *10^{-31} * 1.37 *10^{6}}[/tex]
=> [tex]\lambda = 5.30 *10^{-10 } \ m[/tex]
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Rewritten by : Barada
The de Broglie wavelength of an electron moving at 1.37 x 10⁶ m/s is calculated to be approximately 5.29 x 10⁻¹⁰ meters using de Broglie's equation. Planck's constant and the mass of the electron are used in the formula to find the wavelength.
The de Broglie wavelength of an electron can be calculated using de Broglie's equation:
λ = h / mv
where λ is the wavelength, h is Planck's constant (6.626 x 10⁻³⁴ J·s), m is the electron's mass, and v is the velocity. Given:
Velocity,
v = 1.37 x 10⁶ m/s
Mass, m = 9.11 x 10⁻²⁸ g = 9.11 x 10⁻³¹ kg (convert grams to kilograms)
Using the formula, we have:
λ = 6.626 x 10⁻³⁴ J·s / (9.11 x 10⁻³¹ kg * 1.37 x 10⁶ m/s)
Simplifying gives:
λ ≈ 5.29 x 10⁻¹⁰ meters
Thus, the de Broglie wavelength of the electron is approximately 5.29 x 10⁻¹⁰ meters.