Question

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Answer 1

$\huge\underline{\underline{\bf \orange{Question-}}}$

The de-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state will be ⎯

$\huge\underline{\underline{\bf \orange{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• n for ground state = 1
• Z for hydrogen atom = 1

$\large\underline{\underline{\sf To\:Find:}}$

• de-Broglie Wavelength ${\sf (\lambda)}$

We know that -

de-Broglie Wavelength -

✪$\large{\boxed{\bf \blue{\lambda=\dfrac{h}{p}} }}$

p = mv = momentum

h = Planck's Constant

According to Bohr -

✪$\large{\boxed{\bf \blue{mvr=\dfrac{nh}{2π}} }}$

We can write = mv = p

So ,

$\implies{\sf pr = \dfrac{nh}{2π}}$

$\implies{\sf p = \dfrac{1×h}{2π} }$

$\implies{\sf \pink{p = \dfrac{h}{2π}}}$

On Putting Value of p -

$\implies{\sf \lambda = \dfrac{h}{\dfrac{h}{2πr}} }$

$\implies{\sf \lambda = 2πr }$

We know ,

✪$\large{\boxed{\bf \blue{r=0.53×\dfrac{n^2}{Z}} }}$

$\implies{\sf r = 0.53×\dfrac{1^2}{1}}$

$\implies{\sf \pink{r=0.53A°} }$

On Putting Value of r ⎯

$\implies{\sf \lambda = 2πr}$

$\implies{\sf \lambda = 2×3.14×0.53 }$

$\implies{\bf \red{\lambda = 3.3A°}}$

$\huge\underline{\underline{\bf \orange{Answer-}}}$

Option (2) 3.3A°

The de-Broglie wavelength associated with electrons revolving round the nucleus in a hydrogen atom in ground state will be ${\bf \red{3.3A°}}$.