Question

समान ऊर्जा के एक प्रोटॉन तथा एक अल्फा कण से संबंधित डी ब्रोग्ली तरंग धैर्य का अनुपात क्या होगा ​

Answer 1

Answer:

डी ब्रोगली परिकल्पना समीकरण या डी ब्रोग्ली का सिद्धांत :

डी ब्रोगली के अनुसार प्रकाश विकिरणों के समान अन्य सूक्ष्म कण जैसे electron , प्रोटोन आदि भी द्वेत (dual) प्रकृति दर्शाते है।

Hope it helps you

Answer 2

Given that,

The de Broglie wave endurance corresponding to a proton of equal energy and an alpha particle.

We know that,

de Broglie wavelength :

The wavelength of the particle is inversely proportional to the momentum of the particle.

In mathematically,

$\lambda\proppto\dfrac{1}{mv}$

$\lambda=\dfrac{h}{mv}$

Where, m = mass of particle

v = velocity of particle

h = planck's constant

We need to calculate the wavelength of proton

Using formula of de Broglie wavelength

$\lambda_{p}=\dfrac{h}{m_{p}v}$....(I)

We know that,

The mass of alpha particle is equal to the 2 times of mass of proton .

$m_{\alpha}=2m_{p}$

We need to calculate the wavelength of proton

Using formula of de Broglie wavelength

$\lambda_{\alpha}=\dfrac{h}{m_{\alpha}v}$....(II)

We need to calculate the ratio of wavelength of alpha particle and proton

Divided equation (I) by equation (II)

$\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\dfrac{\dfrac{h}{m_{p}v}}{\dfrac{h}{m_{\alpha}v}}$

Put the value into the formula

$\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\dfrac{2m_{p}}{m_{p}}$

$\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\dfrac{2}{1}$

Hence, The ratio of wavelength of alpha particle and proton is 2:1