Question

If the velocity of electron in Bohr’s first orbit is 6.626 106 m/s, calculate de Broglie wavelength (approximately) associated with it?

Answer 1

Explanation:

Correct option is

C

3.32×10

−10

m

According to de Broglie's equation

λ=

mv

h

where, λ= de-Broglie wavelength

h=6.626×10

−34

Js=Planck's constant

m=mass of electron=9.1×10

−31

kg

velocity of electron, v=2.19×10

6

m/s

upon substitution we get:

λ=

9.1×10

−31

×2.19×10

6

6.626×10

−34

λ=0.332×10

−9

=3.32×10

−10

m

Answer 2

$\large \dag$ Question :-

If the velocity of electron in Bohr’s first orbit is 6.626 × 10⁶ m/s, calculate de Broglie wavelength (approximately) associated with it ?

$\large \dag$ Answer :-

$\red\dashrightarrow\underline{\underline{\sf  \green{Wavelength   \:  is \:  1.1 \: A°}} }$

$\large \dag$ Step by step Explanation :-

❒ We Know that according to de - Broglie equation :-

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{   \blue{  \pmb\lambda= \small\frac{h}{mv} \: \: \: }}}}$

where,

$\large \lambda$ = de - Broglie wavelength

• m = mass of electron

• v = velocity of electron

• h = Planck's Constant

☆ Now Here we have,

$\sf Planck's \: const. = \pmb{h = 6.626 \times {10}^{ - 34} Js}$

$\sf mass \: of \: e {}^{ - } = \pmb{m = 9.1 \times1 {0}^{ - 31} kg }$

$\sf vel. \: of \: e {}^{ - } = \pmb{v =6.626 \times1 {0}^{6} m/s }$

⏩ Putting Values in de - Broglie equation ;

$:\longmapsto \rm \lambda = \frac{6.626 \times 10 {}^{ - 34} }{(9.1 \times {10}^{ - 31})(6.626 \times {10}^{6}) }$

$:\longmapsto \rm \lambda = \frac{\cancel{6.626 } \times 10 {}^{ - 34} }{9.1 \times {10}^{ - 31} \times \cancel{6.626 }\times {10}^{6} }$

$:\longmapsto \rm \lambda = \frac{10 {}^{ - 34} }{9.1 \times {10}^{ - 31} \times {10}^{6} }$

$:\longmapsto \rm \lambda = \frac{ {10}^{ - 34} }{9.1 \times {10}^{ - 25} }$

$:\longmapsto \rm \lambda = \frac{10 {}^{ - 34} \times {10}^{25} }{9.1}$

$:\longmapsto \rm \lambda = \frac{ {10}^{ - 9} }{9.1}$

$:\longmapsto \rm \lambda = 0.109 \times {10}^{ - 9}$

$[/tex]</p><p></p><p>[tex]\purple{ :\longmapsto  \underline {\boxed{{\pmb{ \frak{ \lambda = 1.1 \times {10}^{ - 10 } \: m \: (approx) } }}}}}$

$\blue\rm \large \blue\bigstar \red{ \rm\: In \: Angstorm( A°) : - }$

$\rm \lambda = 1.1 \times {10}^{ - 10 } \times {10}^{10} A°$

$\purple{ \large :\longmapsto  \underline {\boxed{{\pmb{ \frak{ \lambda = \text{1.1 \: A°}\: (approx) } }}}}}$