The de Brogile wavelength of the electron in the ground state of the hydrogen atom is (Radius of the first orbit of hydrogen atom = $0.53 \circ A$

1.67 \circ A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3.33 \circ A

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

106 \circ A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.53 \circ A

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $3.33 \circ A$
$2 π r = n λ \Rightarrow 2 π r = 1 \times λ \Rightarrow λ = 2 π r = 2 \times 3.14 \times 0.53 = 3.33 \circ A$

Suggest Corrections

Similar questions

0.53 A^{\circ}

= 0.53 ˚ A

n

3.4 e V .

The de-brogile wavelength of an electron in the ground state of hydrogen atom is:

$[K . E . = 13.6 e V; 1 e V = 1.602 \times 10^{- 19} J]$

Join BYJU'S Learning Program