Determine the minimum electric potential difference (in volts) that must be applied to accelerate an α-particle so that on interaction with a hydrogen atom, a ground state electron will be excited to n = 6.
E=qV
E=-k((1/nf^2)-(1/ni^2))
V=(-k/q)(1/25-1/1)*(24/25)
V=(2.18e-18/1.602e-19 J)*(24/25)
V=13.1 Volt
13.1 eVolt
To determine the minimum electric potential difference that must be applied to accelerate an α-particle, we need to consider the energy required to excite a hydrogen atom's ground state electron to n = 6.
The energy difference between two energy levels of a hydrogen atom can be calculated using the Rydberg formula:
ΔE = R(1/n₁² - 1/n₂²),
where:
- ΔE is the energy difference,
- R is the Rydberg constant (approximately 2.18 × 10^(-18) J),
- n₁ is the initial energy level (in this case, n = 1 for the hydrogen atom's ground state),
- n₂ is the final energy level (in this case, n = 6).
To find the energy required to excite the electron to n = 6, we will set n₁ = 1 and n₂ = 6 in the formula:
ΔE = R(1/1² - 1/6²)
= R(1 - 1/36)
= R(35/36).
Now, we need to convert the energy difference (ΔE) into electric potential difference (V) using the formula:
ΔE = qV,
where:
- q is the charge of the α-particle,
- V is the electric potential difference.
The charge of an α-particle is twice the charge of a proton, which is 2 × 1.6 × 10^(-19) C.
Setting ΔE = qV, we can solve for V:
R(35/36) = (2 × 1.6 × 10^(-19) C) × V.
Simplifying the equation, we can solve it for V:
V = (R(35/36))/(2 × 1.6 × 10^(-19) C).
Now, we can substitute the values of R and solve for V:
V = (2.18 × 10^(-18) J × (35/36))/(2 × 1.6 × 10^(-19) C).
Calculating this expression will give us the minimum electric potential difference required to accelerate the α-particle so that the electron of a hydrogen atom is excited to n = 6.