The wavelength of electrons accelerated from rest through a potential difference of 40 kV is \( x \times 10^{-12} \text{ m} \). The value of \( x \) is _________. (Nearest integer)
Given Data:
- Mass of electron (\(m\)) = \( 9.1 \times 10^{-31} \text{ kg} \)
- Charge on an electron (\(e\)) = \( 1.6 \times 10^{-19} \text{ C} \)
- Planck's constant (\(h\)) = \( 6.63 \times 10^{-34} \text{ J s} \)
Detailed Step-by-Step Solution
This problem is based on the De Broglie Wavelength of a charged particle accelerated through a potential difference.
Step 1: Write Down the Relevant Formula
The kinetic energy (\(K\)) acquired by an electron accelerated through a potential difference (\(V\)) is \( K = eV \).
The De Broglie wavelength is given by:
Step 2: Prepare the Values
- \( V = 40 \text{ kV} = 40 \times 10^3 \text{ V} = 4 \times 10^4 \text{ V} \)
- \( m = 9.1 \times 10^{-31} \text{ kg} \)
- \( e = 1.6 \times 10^{-19} \text{ C} \)
- \( h = 6.63 \times 10^{-34} \text{ J s} \)
Step 3: Calculate the Denominator Term (\(\sqrt{2meV}\))
First, calculate the value inside the square root (\(2meV\)):
\( 2meV = (2 \times 9.1 \times 1.6 \times 4) \times 10^{-31 - 19 + 4} \)
\( 2meV = 116.48 \times 10^{-46} \)
Now, take the square root:
\( \sqrt{2meV} \approx 10.792 \times 10^{-23} \text{ kg m/s} \)
Step 4: Calculate Wavelength (\(\lambda\))
Substitute the denominator back into the main formula:
\( \lambda = \left( \frac{6.63}{10.792} \right) \times 10^{-34 - (-23)} \)
\( \lambda \approx 0.6143 \times 10^{-11} \text{ m} \)
We need the answer in the format \( x \times 10^{-12} \text{ m} \):
Conclusion: Comparing our result \( 6.143 \times 10^{-12} \text{ m} \) with \( x \times 10^{-12} \text{ m} \), we find that \( x = 6.143 \). Rounding to the nearest integer, the value of \( x \) is 6.
Shortcut Check: Using \( \lambda = \frac{12.27}{\sqrt{V}} \text{ \AA} \), we get \( \lambda = \frac{12.27}{\sqrt{40000}} = \frac{12.27}{200} = 0.06135 \text{ \AA} = 6.135 \times 10^{-12} \text{ m} \), which confirms our integer is 6.
Acing De Broglie Wavelength Calculations
This numerical is a staple in JEE Main and NEET. While it's great to know the full calculation to avoid errors when non-standard masses or charges are given, memorizing the shortcut formula for electrons (\( \lambda = \sqrt{150/V} \text{ \AA} \) or \( \lambda = 12.27/\sqrt{V} \text{ \AA} \)) saves immense amounts of time during the exam.
To master both the long-form derivations and the time-saving shortcuts for quantum mechanical calculations, review our complete notes on the Structure of Atom Class 11 Chemistry.
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