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Step-by-Step Solution
Step 1: Write down the de Broglie wavelength formula
The wavelength \lambda of an electron accelerated from rest through a potential difference V is given by:
\lambda = \frac{h}{\sqrt{2 m q V}}
where:
• h is Planck’s constant,
• m is the mass of the electron,
• q is the charge on the electron,
• V is the accelerating potential difference.
Step 2: Substitute the known values
The given constants are:
• m = 9.1 \times 10^{-31}\,\text{kg}
• q = 1.6 \times 10^{-19}\,\text{C}
• V = 40\,\text{kV} = 40 \times 10^3\,\text{V}
• h = 6.63 \times 10^{-34}\,\text{J·s}
Substituting into the formula:
\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40 \times 10^3}}
Step 3: Simplify under the square root
Inside the square root, calculate the product step by step:
2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40 \times 10^3
Combine the numerical factors first, then handle the power-of-ten parts:
Numerically:
• 2 \times 9.1 = 18.2
• 18.2 \times 1.6 = 29.12
• 29.12 \times 40 = 1164.8
Exponentially:
• 10^{-31} \times 10^{-19} \times 10^3 = 10^{-31-19+3} = 10^{-47}
Hence, inside the square root:
1164.8 \times 10^{-47}
Thus, the denominator is:
\sqrt{1164.8 \times 10^{-47}}
Step 4: Final calculation of \lambda
Divide the numerator by the square root above:
\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{1164.8 \times 10^{-47}}}
Numerically, this approximates to:
6.144 \times 10^{-12}\,\text{m} \approx 6 \times 10^{-12}\,\text{m}
Step 5: Identify the value of x
The question states \lambda = x \times 10^{-12}\,\text{m} and the nearest integer to x is 6.
Therefore,
x = 6.
