Answer
425.1k+ views
Hint: The de Broglie wavelength is calculated by dividing the Planck's constant with the momentum of the electron. The radius of the nth orbital is equal to the square of n multiplied by the radius of the first orbit.
Complete answer: Let us understand the de Broglie wavelength:
The wavelength of the wave associated with any material particle was calculated by analogy with photons as follows:
In the case of a photon, if it is assumed to have wave character, its energy is given by,
\[E=h\text{v}\]
Where v is the frequency of the wave and h is the planck's constant.
If the photon is supposed to have a particle nature, its energy is,
\[\text{E=}m{{c}^{2}}\]
Where m is the mass of the particle and c is the speed of light.
On solving both the equation, we get
\[h\text{v}=m{{c}^{2}}\]
\[but\text{ v=}\dfrac{c}{\lambda }\]
\[h\dfrac{c}{\lambda }=m{{c}^{2}}\text{ or }\lambda =\dfrac{h}{mc}\]
For applying it on any material, the c is replaced with v (velocity of the material).
$\lambda =\dfrac{h}{m\text{v}}$ .......(i)
This is called the de Broglie equation and \[\lambda \]is the de Broglie wavelength.
The radii of the ground states of the hydrogen atom can be expressed by,
\[{{r}_{n}}={{n}^{2}}{{r}_{0}}\]
Where\[{{r}_{0}}\] is the radius of the first stationary state and is called the Bohr radius.
Now, according to question the radius of the 3rd orbit is equal to
\[{{r}_{3}}={{3}^{2}}x\]
Given in the question is ‘x’. Hence the radius of 3rd orbit is
\[{{r}_{3}}=9x\]
Now, according to Bohr’s postulate of angular momentum,
\[mvr=n\dfrac{h}{2\pi }\]
Putting the values in the above equation, we get
\[mv9x=3\dfrac{h}{2\pi }\]
On solving the above equation,
\[\dfrac{2\pi \text{ x }9x}{3}=\dfrac{h}{m\text{v}}\]
\[\dfrac{h}{m\text{v}}=6\pi x\]
$\lambda =6\pi x$ .........[From equation (i)]
Hence the de Broglie wavelength is \[6\pi x\]
So, the correct answer is “Option B”.
Note: All the formulas and equations should be used correctly. The significance of the de Broglie equation lies in the fact that it relates to the particle character with the wave character of matter. The de Broglie equation applies to all objects.
Complete answer: Let us understand the de Broglie wavelength:
The wavelength of the wave associated with any material particle was calculated by analogy with photons as follows:
In the case of a photon, if it is assumed to have wave character, its energy is given by,
\[E=h\text{v}\]
Where v is the frequency of the wave and h is the planck's constant.
If the photon is supposed to have a particle nature, its energy is,
\[\text{E=}m{{c}^{2}}\]
Where m is the mass of the particle and c is the speed of light.
On solving both the equation, we get
\[h\text{v}=m{{c}^{2}}\]
\[but\text{ v=}\dfrac{c}{\lambda }\]
\[h\dfrac{c}{\lambda }=m{{c}^{2}}\text{ or }\lambda =\dfrac{h}{mc}\]
For applying it on any material, the c is replaced with v (velocity of the material).
$\lambda =\dfrac{h}{m\text{v}}$ .......(i)
This is called the de Broglie equation and \[\lambda \]is the de Broglie wavelength.
The radii of the ground states of the hydrogen atom can be expressed by,
\[{{r}_{n}}={{n}^{2}}{{r}_{0}}\]
Where\[{{r}_{0}}\] is the radius of the first stationary state and is called the Bohr radius.
Now, according to question the radius of the 3rd orbit is equal to
\[{{r}_{3}}={{3}^{2}}x\]
Given in the question is ‘x’. Hence the radius of 3rd orbit is
\[{{r}_{3}}=9x\]
Now, according to Bohr’s postulate of angular momentum,
\[mvr=n\dfrac{h}{2\pi }\]
Putting the values in the above equation, we get
\[mv9x=3\dfrac{h}{2\pi }\]
On solving the above equation,
\[\dfrac{2\pi \text{ x }9x}{3}=\dfrac{h}{m\text{v}}\]
\[\dfrac{h}{m\text{v}}=6\pi x\]
$\lambda =6\pi x$ .........[From equation (i)]
Hence the de Broglie wavelength is \[6\pi x\]
So, the correct answer is “Option B”.
Note: All the formulas and equations should be used correctly. The significance of the de Broglie equation lies in the fact that it relates to the particle character with the wave character of matter. The de Broglie equation applies to all objects.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)