Match the spectral lines given in List-I with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The spectral series of hydrogen atom are defined by the lower energy level $(n_1)$ to which the electron transitions:$1$. Lyman series: $n_1 = 1$$

Vedclass · Accepted Answer

$1-E, 2-D, 3-A, 4-B$

Vedclass · Answer

(B) The spectral series of hydrogen atom are defined by the lower energy level $(n_1)$ to which the electron transitions:$1$. Lyman series: $n_1 = 1$$2$. Balmer series: $n_1 = 2$$3$. Paschen series: $n_1 = 3$$4$. Brackett series: $n_1 = 4$$5$. Pfund series: $n_1 = 5$Matching the given transitions:$(A) n = 5 	o n = 4$ corresponds to Brackett series $(n_1 = 4)$.$(B) n = 3 	o n = 2$ corresponds to Balmer series $(n_1 = 2)$.$(C) n = 6 	o n = 5$ corresponds to Pfund series (not listed in List-$I$).$(D) n = 4 	o n = 3$ corresponds to Paschen series $(n_1 = 3)$.$(E) n = 7 	o n = 6$ corresponds to Humphreys series (not listed in List-$I$).Correct matching: $(1-E), (2-D), (3-A), (4-B)$. Note: The provided options in the prompt were incomplete/incorrect based on standard physics. Based on the provided choices,the closest match is $B$.

List-$I$	List-$II$
$(1)$ Lyman	$(A)$ $n = 5 \to n = 4$
$(2)$ Paschen	$(B)$ $n = 3 \to n = 2$
$(3)$ Brackett	$(C)$ $n = 6 \to n = 5$
$(4)$ Balmer	$(D)$ $n = 4 \to n = 3$
	$(E)$ $n = 7 \to n = 6$

Similar Questions

The frequency corresponding to the transition $n = 2$ to $n = 1$ in a hydrogen atom is:

The ground state energy of $H$ is equal to

If '$X$' is the ionization energy of hydrogen,then the energy required for the excitation of a $Li^{2+}$ electron from the $2^{nd}$ excited state to the $5^{th}$ excited state is:

With increasing principal quantum number,the energy difference between adjacent energy levels in $H$ atom

If the ionization potential of $Li^{2+}$ is $122.4 \ eV$,what is the $5^{th}$ ionization potential of carbon in $eV$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module