हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान कक्षा ११

A Proton and an Electron Are Accelerated by the Same Potential Difference. Let λE and λPdenote the De Broglie Wavelengths of the Electron and the Proton, Respectively.

Advertisements
Advertisements

प्रश्न

A proton and an electron are accelerated by the same potential difference. Let λe and λpdenote the de Broglie wavelengths of the electron and the proton, respectively.

विकल्प

  • λe = λp

  • λe < λp

  • λe > λp

  • The relation between λe and λp depends on the accelerating potential difference.

MCQ
Advertisements

उत्तर

λe > λp

Let me and mp be the masses of electron and proton, respectively.
Let the applied potential difference be V.

Thus, the de-Broglie wavelength of the electron,

`λ_e = h/sqrt(2m_eeV)`    ....(1)

And de-Broglie wavelength of the proton,

`λ_p = h/sqrt(2m_peV)`        ....(2)

Dividing equation (2) by equation (1), we get :

`λ_p/λ_e = sqrt(m_e)/sqrt(m_p)`

`m_e < m_p`

`therefore λ_p/λ_e < 1`

⇒ `λ_p < λ_e`

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 42: Photoelectric Effect and Wave-Particle Duality - MCQ [पृष्ठ २६४]

APPEARS IN

एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 42 Photoelectric Effect and Wave-Particle Duality
MCQ | Q 14 | पृष्ठ २६४
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×