According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is 10-9 m . The principle quantum number for this electron is

Question

According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is 10-9 m . The principle quantum number for this electron is (A) 1 (B) 2 (C) 3 (D) 4

Tardigrade Admin · Accepted Answer

2 π r n=λ ⇒ n=(λ/2 π r) =(10-9/2 × 3.14 × 5.13 × 10-11)=3

Q. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is $1 0^{- 9} m .$ The principle quantum number for this electron is

1

2

3

4

$2 π r n = λ \Rightarrow n = \frac{λ}{2 π r}$
$= \frac{1 0 ^{- 9}}{2 \times 3.14 \times 5.13 \times 1 0 ^{- 11}} = 3$

Q. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is 10−9m. The principle quantum number for this electron is

1

2

3

4

2πrn=λ⇒n=2πrλ​ =2×3.14×5.13×10−1110−9​=3

Q. According to de-Broglie, the de-Broglie wavelength for electron in an orbit of hydrogen atom is $1 0^{- 9} m .$ The principle quantum number for this electron is

$2 π r n = λ \Rightarrow n = \frac{λ}{2 π r}$
$= \frac{1 0 ^{- 9}}{2 \times 3.14 \times 5.13 \times 1 0 ^{- 11}} = 3$