If the velocity of the electron in Bohr’s first orbit is 2.19 × 10^6 ms^–1, calculate the de Broglie wavelength associated with it.
Ncert solutions for Class 11th Chemistry Part - 1 Chapter 2 Exercise 59
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Answered by
192
v = 2.19 * 10⁶ m/s
λ = h / p
h = Planck's constant = 6.636 * 10⁻³⁴ units
m = mass of an electron = 9.11 * 10⁻³¹ kg
Bohr's first orbit : n = 1, K shell in an atom... like Hydrogen atom..
λ = 6.636 * 10⁻³⁴ / [ 9.11 * 10⁻³¹ * 2.19 * 10⁶ ] meters
= 0.3326 * 10⁻⁹ meters
λ = h / p
h = Planck's constant = 6.636 * 10⁻³⁴ units
m = mass of an electron = 9.11 * 10⁻³¹ kg
Bohr's first orbit : n = 1, K shell in an atom... like Hydrogen atom..
λ = 6.636 * 10⁻³⁴ / [ 9.11 * 10⁻³¹ * 2.19 * 10⁶ ] meters
= 0.3326 * 10⁻⁹ meters
Answered by
64
Answer:
The de Broglie wavelength associated with it is 3.311 angstrom.Explanation:
De-Broglie wavelength is calculated by using the formula:
where,
= wavelength of electron
h = Planck's constant =
m = mass of electron =
v = velocity of electron =
The de Broglie wavelength associated with it is 3.311 angstrom.
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