Answer :
The accelerating potential needed to produce electrons of wavelength 6.00 nm is 0.0415 volts.
Using the de Broglie wavelength formula, we can find the momentum of the electron and then the accelerating potential. as,
λ = h/p
∴ p = h/λ = 6.6 × 10⁻³⁴/6 × 10⁻⁹ = 1.1 × 10⁻²⁵ Kg m/s.
The momentum of an electron can be expressed in terms of its kinetic energy (K) as:
[tex]p=\sqrt{2mK}[/tex] (where m is the mass of the electron)
And we know, the kinetic energy of the electron as,
K = eV (where e is the elementary charge)
∴ [tex]p=\sqrt{2meV}[/tex]
∴ [tex]V=\frac{p^{2} }{2me}[/tex]
Now, substituting the values of momentum, mass and charge;
we get:
V = (1.1 × 10⁻²⁵)² / (2 * 9.1 x 10⁻³¹ kg * 1.6 x 10⁻¹⁹ C)
= 0.0415 V
Therefore, the accelerating potential needed to produce electrons of wavelength 6.00 nm is 0.0415 V (or, 41.5 mV).
Learn more about de Broglie equation here
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