Question:
Flame photometry is based on measurement of
Flame photometry is based on measurement of
Show Hint
Flame photometry is an analytical technique used mainly to measure elements like sodium, potassium, calcium and lithium. The question is about the basic principle — what exactly the instrument measures.
Updated On: Jun 24, 2026
- Absorbed radiation
- Scattered radiation
- Emitted radiation
- Reflected radiation
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Flame photometry is an analytical technique used mainly to measure elements like sodium, potassium, calcium and lithium. The question is about the basic principle — what exactly the instrument measures.
Step 1: In flame photometry, the sample solution is sprayed into a flame. The heat of the flame excites the metal atoms, pushing their electrons to higher energy levels.
Step 2: These excited electrons are unstable and quickly fall back to their lower energy levels. As they do, they release the extra energy as light of a characteristic colour/wavelength. This is emission of radiation, and the brightness of the emitted light is proportional to the amount of the element present.
Step 3: So flame photometry is based on measuring emitted radiation. This is different from atomic absorption spectroscopy, which measures absorbed radiation instead. Scattering and reflection are not the basis of this method.
Answer: Option (3) — Emitted radiation.
Flame photometry is an analytical technique used mainly to measure elements like sodium, potassium, calcium and lithium. The question is about the basic principle — what exactly the instrument measures.
Step 1: In flame photometry, the sample solution is sprayed into a flame. The heat of the flame excites the metal atoms, pushing their electrons to higher energy levels.
Step 2: These excited electrons are unstable and quickly fall back to their lower energy levels. As they do, they release the extra energy as light of a characteristic colour/wavelength. This is emission of radiation, and the brightness of the emitted light is proportional to the amount of the element present.
Step 3: So flame photometry is based on measuring emitted radiation. This is different from atomic absorption spectroscopy, which measures absorbed radiation instead. Scattering and reflection are not the basis of this method.
Answer: Option (3) — Emitted radiation.
Was this answer helpful?
0
0
Top TS PGECET Pharmacy Questions
- Increase in the solubility of the aqueous systems by adding large amount of additives is known as.........
- Clark's formula………
- One of the following is used as a key ingredient in hair conditioning product
- Empty capsule has moisture content in the range of ………
- Aerosol container must withstand pressures as high as 140–180 psi at ………
View More Questions
Top TS PGECET Pharmaceutical Analysis Questions
- If a researcher mistakenly concludes that there is a significant effect when, in fact, there is none, what type of error is committed?
- ……………. is the measure of how close a measurement comes to the true value
- How many vibrational modes are possible for H\textsubscript{2}O?
- Which of the following is the disadvantage of nitrogen, which can be used as carrier gas in gas chromatography?
- What is the nature of mobile phase in normal phase chromatography?
View More Questions
Top TS PGECET Questions
- If \( P = \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{pmatrix} \) is the modal matrix of \( A = \begin{pmatrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{pmatrix} \) then the sum of all the elements of \( P^{-1}AP \) is
- If the system of equations \( kx + y + z = k - 1 \), \( x + ky + z = k - 1 \), and \( x + y + kz = k - 1 \) has infinite solutions, then \( k \) is
- The value of the integral \( \int_C (2xy - x^2) \, dx + (x^2 + y^2) \, dy \) where \( C \) is the boundary of the region enclosed by \( y = x^2 \) and \( y^2 = x \), described in the positive sense, is
- Let \( f(x) = \log x \). The number \( C \) strictly between \( e^2 \) and \( e^3 \) such that its reciprocal is equal to \( \frac{f(e^3) - f(e^2)}{e^3 - e^2} \) is
- The solution of the differential equation \( (D^2 + 2)y = x^2 \) is
View More Questions