The de-broglie wavelength of the electron in the ground state of hydrogen atom is
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The velocity v of the hydrogen electron in its ground state is v = αc where α is the fine structure constant. Hence its momentum is mαc where m is the electron mass. So the de Broglie wavelength is h/mαc = 0.332 nm, same result Tusharkanta Srichandan got.
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So the de Broglie wavelength is h/mαc = 0.332 nm, same result Tusharkanta Srichandan got. (Although I didn't need the given KE of the electron, using the same numbers I used tacitly above for electron mass m and velocity v, its KE is ½mv² = 13.6 ev as it turns out.
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