NEET 2025 | Structure of Atom Question 1 | Chemistry

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes $n=2 \rightarrow n=3$ and $n=4 \rightarrow$ $\mathrm{n}=6$ transitions, respectively, is

$\frac{1}{9}$

$\frac{1}{4}$

$\frac{1}{36}$

$\frac{1}{16}$

NEET 2025

MCQ (Single Correct Answer)

-1

Energy and radius of first Bohr orbit of $\mathrm{He}^{+}$and $\mathrm{Li}^{2+}$ are [Given $\mathrm{R}_{\mathrm{H}}=2.18 \times 10^{-18} \mathrm{~J}, \mathrm{a}_0=52.9 \mathrm{pm}$ ]

$$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}$$

$$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-16} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}$$

$$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=17.6 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=26.4 \mathrm{pm} \end{aligned}$$

$$\begin{aligned} & \mathrm{E}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=-8.72 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{Li}^{2+}\right)=26.4 \mathrm{pm} \\ & \mathrm{E}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=-19.62 \times 10^{-18} \mathrm{~J} ; \\ & \mathrm{r}_{\mathrm{n}}\left(\mathrm{He}^{+}\right)=17.6 \mathrm{pm} \end{aligned}$$

NEET 2024 (Re-Examination)

MCQ (Single Correct Answer)

-1

The quantum numbers of four electrons are given below :

I. $$n=4 ; I=2 ; m_1=-2 ; s=-\frac{1}{2}$$

II. $$n=3 ; I=2 ; m_1=1 ; s=+\frac{1}{2}$$

III. $$n=4 ; I=1 ; m_1=0 ; s=+\frac{1}{2}$$

IV. $$n=3 ; I=1 ; m_1=-1 ; s=+\frac{1}{2}$$

The correct decreasing order of energy of these electrons is

IV $$>$$ II $$>$$ III $$>$$ I

I $$>$$ III $$>$$ II $$>$$ IV

III $$>$$ I $$>$$ II $$>$$ IV

I $$>$$ II $$>$$ III $$>$$ IV

NEET 2024 (Re-Examination)

MCQ (Single Correct Answer)

-1

Given below are two statements:

Statement I: The Balmer spectral line for H atom with lowest energy is located at $$\frac{5}{36} R_H \mathrm{~cm}^{-1}$$.

($$\mathrm{R}_{\mathrm{H}}=$$ Rydberg constant)

Statement II: When the temperature of blackbody increases, the maxima of the curve (intensity and wavelength) shifts to shorter wavelength.

In the light of the above statements, choose the correct answer from the options given below:

Statement I is true but Statement II is false

Statement I is false but Statement II is true

Both Statement I and Statement II are true

Both Statement I and Statement II are false

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