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What is the wavelength of the electron in the ground state of the hydrogen atom if its velocity is 1.70x10^6 ?

Answer :

Final answer:

The wavelength of electron in the ground state of the hydrogen atom with a velocity of 1.70x10^6 m/s is approximately 6.55 x 10^-10 meters.

Explanation:

To calculate the wavelength of the electron in the ground state of the hydrogen atom, we can use the de Broglie wavelength formula:

λ = h / p

Where:

λ = wavelength

h = Planck's constant (6.626 x 10^-34 J·s)

p = momentum

First, we need to calculate the momentum of the electron, which can be calculated using the equation:

p = m * v

Where:

p = momentum

m = mass of the electron (9.109 x 10^-31 kg)

v = velocity (1.70 x 10^6 m/s)

Now, calculate the momentum:

p = (9.109 x 10^-31 kg) * (1.70 x 10^6 m/s) ≈ 1.55 x 10^-24 kg·m/s

Now, we can calculate the wavelength using the de Broglie formula:

λ = (6.626 x 10^-34 J·s) / (1.55 x 10^-24 kg·m/s) ≈ 6.55 x 10^-10 meters

So, the wavelength of the electron in the ground state of the hydrogen atom with a velocity of 1.70x10^6 m/s is approximately 6.55 x 10^-10 meters.

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