According to Bohr's quantization condition (postulate), the angular momentum of an electron in a stationary orbit is quantized:
$$mvr_n = n\frac{h}{2\pi}$$
where $m$ is the electron mass, $v$ is its velocity, $r_n$ is the orbit radius, $h$ is Planck's constant, and $n$ is the principal quantum number.
According to the de Broglie hypothesis, the wavelength ($\lambda$) associated with the electron is:
$$\lambda = \frac{h}{mv}$$
Rearranging this gives:
$$mv = \frac{h}{\lambda}$$
Substitute the de Broglie relation for $mv$ into the Bohr quantization condition:
$$\left(\frac{h}{\lambda}\right) r_n = n\frac{h}{2\pi}$$
Simplifying the equation by canceling $h$ on both sides, we get the fundamental relationship between the orbit radius ($r_n$) and the de Broglie wavelength ($\lambda$):
$$\frac{r_n}{\lambda} = \frac{n}{2\pi}$$
$$2\pi r_n = n\lambda$$
We are given the following values and formula:
de Broglie wavelength: $\lambda = 13.4 \text{ Å}$
Radius of $n^{\text{th}}$ Bohr orbit: $r_n = 0.53n^2 \text{ Å}$
Value of $\pi = 3.14$
Substitute the expressions for $r_n$ and $\lambda$ into the derived equation $2\pi r_n = n\lambda$:
$$2(3.14) (0.53n^2 \text{ Å}) = n (13.4 \text{ Å})$$
First, simplify the constants on the left side:
$$(2 \times 3.14 \times 0.53) n^2 = 13.4 n$$
$$3.3392 n^2 = 13.4 n$$
Since $n$ is the principal quantum number, it cannot be zero. We can divide both sides by $n$:
$$3.3392 n = 13.4$$
$$n = \frac{13.4}{3.3392}$$
$$n \approx 4.0128$$
The question asks for the value of $n$ rounded up to the nearest integer.
$$n \approx 4.0128$$
Rounding $4.0128$ to the nearest integer gives:
$$n = 4$$
The value of $n$ is 4.
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