Question

The speed of electrons in a scanning electron microscope is \(1 \times 10^7\) ms⁻¹. If the protons having the same speed are used instead of electrons, then the resolving power of scanning proton microscope will be changed by a factor of:

• A. 1837
• B. \(\frac{1}{1837}\)
• C. \(\sqrt{1837}\)
• D. \(\frac{1}{\sqrt{1837}}\)

Hint:

Heavier particles have smaller de-Broglie wavelengths at the same speed, leading to much higher resolution.

Answer 1

The Correct Option is A

To determine the change in the resolving power of a scanning microscope when protons are used instead of electrons, we need to consider the de Broglie wavelength relation. The resolving power of a microscope is inversely proportional to the wavelength of the particles used.

The de Broglie wavelength \(\lambda\) is given by: 

\(\lambda = \frac{h}{mv}\)

where:

• \(h\) is Planck's constant
• \(m\) is the mass of the particle
• \(v\) is the speed of the particle

For electrons and protons at the same speed, the formula becomes:

\(\lambda_e = \frac{h}{m_e v}\) and \(\lambda_p = \frac{h}{m_p v}\)

where:

• \(\lambda_e\) is the wavelength of electrons
• \(\lambda_p\) is the wavelength of protons
• \(m_e\) is the mass of an electron
• \(m_p\) is the mass of a proton

 

The ratio of their wavelengths is:

\(\frac{\lambda_p}{\lambda_e} = \frac{m_e}{m_p}\)

Given that the mass of a proton is approximately 1837 times that of an electron \((m_p = 1837 \cdot m_e)\), the ratio simplifies to:

\(\frac{\lambda_p}{\lambda_e} = \frac{1}{1837}\)

As the resolving power \((RP)\) is inversely proportional to the wavelength:

\(\text{RP} \propto \frac{1}{\lambda}\)

The resolving power of the scanning proton microscope compared to the electron microscope is given by:

\(\frac{\text{RP}_p}{\text{RP}_e} = \frac{\lambda_e}{\lambda_p} = 1837\)

Thus, the resolving power changes by a factor of 1837. Therefore, the correct answer is:

1837