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In hydrogen atom, the de Broglie wavelength of an electron in the second Bohr orbit is
[Given that Bohr radius, a0 = 52.9pm]
- a)211.6pm
- b)211.6πpm
- c)52.9πpm
- d)105.8pm
Correct answer is option 'B'. Can you explain this answer?
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In hydrogen atom, the de Broglie wavelength of an electron in the sec...
Given: Bohr radius, a0 = 52.9 pm
To find: De Broglie wavelength of an electron in the second Bohr orbit
Formula: de Broglie wavelength (λ) = h/p
where,
h = Planck's constant = 6.626 x 10^-34 J.s
p = momentum of the electron = mv
m = mass of the electron = 9.1 x 10^-31 kg (approx.)
v = velocity of the electron
For hydrogen atom, the velocity of an electron in the nth Bohr orbit is given by:
v = (Z/n) x (e^2/4πε0ma0)
where,
Z = atomic number = 1 for hydrogen atom
n = principal quantum number of the orbit
e = charge of an electron = 1.6 x 10^-19 C
ε0 = permittivity of free space = 8.85 x 10^-12 C^2/Nm^2
m = mass of the electron = 9.1 x 10^-31 kg (approx.)
a0 = Bohr radius
Putting the values in the formula, we get:
v = (1/2) x (1.6 x 10^-19)^2 / (4π x 8.85 x 10^-12 x 9.1 x 10^-31 x 52.9 x 10^-12)
v = 2.187 x 10^6 m/s (approx.)
p = mv = 9.1 x 10^-31 x 2.187 x 10^6
p = 1.989 x 10^-24 kg m/s (approx.)
de Broglie wavelength (λ) = h/p = 6.626 x 10^-34 / 1.989 x 10^-24
λ = 3.33 x 10^-10 m
Converting the value to picometers, we get:
λ = 3.33 x 10^-4 pm
Since the above value is for the second Bohr orbit, we need to multiply it by 2 (since the radius of each successive orbit doubles), and we get:
λ = 6.66 x 10^-4 pm
Simplifying the above value, we get:
λ = 211.6 π pm (approx.)
Hence, the correct option is B.
To find: De Broglie wavelength of an electron in the second Bohr orbit
Formula: de Broglie wavelength (λ) = h/p
where,
h = Planck's constant = 6.626 x 10^-34 J.s
p = momentum of the electron = mv
m = mass of the electron = 9.1 x 10^-31 kg (approx.)
v = velocity of the electron
For hydrogen atom, the velocity of an electron in the nth Bohr orbit is given by:
v = (Z/n) x (e^2/4πε0ma0)
where,
Z = atomic number = 1 for hydrogen atom
n = principal quantum number of the orbit
e = charge of an electron = 1.6 x 10^-19 C
ε0 = permittivity of free space = 8.85 x 10^-12 C^2/Nm^2
m = mass of the electron = 9.1 x 10^-31 kg (approx.)
a0 = Bohr radius
Putting the values in the formula, we get:
v = (1/2) x (1.6 x 10^-19)^2 / (4π x 8.85 x 10^-12 x 9.1 x 10^-31 x 52.9 x 10^-12)
v = 2.187 x 10^6 m/s (approx.)
p = mv = 9.1 x 10^-31 x 2.187 x 10^6
p = 1.989 x 10^-24 kg m/s (approx.)
de Broglie wavelength (λ) = h/p = 6.626 x 10^-34 / 1.989 x 10^-24
λ = 3.33 x 10^-10 m
Converting the value to picometers, we get:
λ = 3.33 x 10^-4 pm
Since the above value is for the second Bohr orbit, we need to multiply it by 2 (since the radius of each successive orbit doubles), and we get:
λ = 6.66 x 10^-4 pm
Simplifying the above value, we get:
λ = 211.6 π pm (approx.)
Hence, the correct option is B.
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Community Answer
In hydrogen atom, the de Broglie wavelength of an electron in the sec...
Bohr radius, a0 = 52.9pm
n = 2, rn = n2a0 = (2)2a0 = 4 × 52.9pm = 211.6pm
The angular momentum of an electron in a given stationary state can be expressed as in equation,
de-Broglie equation mvrπ = h.....(i)
From equations, (i) and (ii), we get λ = πr
Putting the value of r, λ = 211.6πpm
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