

Given :


We, have to find frequency :





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Answer:
a) n = 9.9 b) E₁₀ = 19.25 eV
Explanation:
Solving the Scrodinger equation for the electronegative box we get
Eₙ = (h² / 8m L²2) n²
where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number
In this case En = 19 eV let us reduce to the SI system
En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J
n = √ (In 8 m L² / h²)
let's calculate
n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²
n = √ (98) n = 9.9
since n must be an integer, we approximate them to 10
b) We substitute for the calculation of energy
In = (h² / 8mL2² n²
In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²
E₁₀ = 3.08 10⁻¹⁸ J
we reduce eV
E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)
E₁₀ = 1.925 101 eV
E₁₀ = 19.25 eV
the result with significant figures is
E₁₀ = 19.25 eV
Answer: You didn't put what a, b, c, or d is so how is anyone supposed to know what the answer is?
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;

Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.


f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;

Here, p is the power of the lens.
Put f= 35.71 cm.

p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
1 and 4 are tire.
2 and 3 are not.