Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Structure of Atom question

2022 · 28 Jun · Shift 1 · Q15
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Structure of Atom
  5. /2022 · 28 Jun · Shift 1 · Q15

Structure of Atom question

2022 · 28 Jun · Shift 1 · Q15

JEE MainChemistryStructure of AtomNumerical+4 / −1
If the work function of a metal is 6.63 ×\times× 10 −-− 19J, the maximum wavelength of the photon required to remove a photoelectron from the metal is ‾\underline{\hspace{2cm}}​ nm. (Nearest integer) [Given : h = 6.63 ×\times× 10 −-− 34 J s, and c = 3 ×\times× 108 m s −-− 1]
Numerical answer
View written solutionFree

Correct answer: 300

  1. For photoelectric emission, the threshold photon energy must equal the work function:

E=ϕ=hcλmax⁡E = \phi = \frac{hc}{\lambda_{\max}}E=ϕ=λmax​hc​

Here,

  • ϕ=6.63×10−19 J\phi = 6.63 \times 10^{-19}\,\text{J}ϕ=6.63×10−19J
  • h=6.63×10−34 J sh = 6.63 \times 10^{-34}\,\text{J s}h=6.63×10−34J s
  • c=3×108 m s−1c = 3 \times 10^8\,\text{m s}^{-1}c=3×108m s−1
  1. Rearranging for maximum wavelength:

λmax⁡=hcϕ\lambda_{\max} = \frac{hc}{\phi}λmax​=ϕhc​

  1. Substitute the values:

λmax⁡=(6.63×10−34)(3×108)6.63×10−19\lambda_{\max} = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{6.63 \times 10^{-19}}λmax​=6.63×10−19(6.63×10−34)(3×108)​

  1. Simplify:

λmax⁡=3×10−2610−19=3×10−7 m\lambda_{\max} = \frac{3 \times 10^{-26}}{10^{-19}} = 3 \times 10^{-7}\,\text{m}λmax​=10−193×10−26​=3×10−7m

More explicitly,

λmax⁡=3×10−7 m\lambda_{\max} = 3 \times 10^{-7}\,\text{m}λmax​=3×10−7m

  1. Convert into nm:

1 m=109 nm1\,\text{m} = 10^9\,\text{nm}1m=109nm

So,

3×10−7 m=3×10−7×109 nm=3×102 nm=300 nm3 \times 10^{-7}\,\text{m} = 3 \times 10^{-7} \times 10^9\,\text{nm} = 3 \times 10^2\,\text{nm} = 300\,\text{nm}3×10−7m=3×10−7×109nm=3×102nm=300nm

  1. Therefore, the nearest integer is:

300\boxed{300}300​

PreviousNext

More from Structure of Atom

  • Consider the following statements : (A) The principal quantum number 'n' is a positive integer with values of 'n' = 1, 2, 3, ... (B) The azimuthal quantum number 'l' for a given 'n' (principal quantum number) can have values as 'l' = 0, 1,…2022 · MCQ
  • The minimum uncertainty in the speed of an electron in an one dimensional region of length 2ao​(Where ao​= Bohr radius 52.9pm) is ​kms−1…2022 · Numerical
  • Given below are the quantum numbers for 4 electrons. A. n=3,l=2, m1​=1, ms​=+1/2 B. n=4,l=1, m1​=0, ms​=+1/2 C. n=4,l=2, m1​=−2, ms​=−1/2…2022 · MCQ
  • Which of the following statements are correct? (A) The electronic configuration of Cr is [Ar] 3d5 4s1. (B) The magnetic quantum number may have a negative value. (C) In the ground state of an atom, the orbitals are filled in order of their…2022 · MCQ
  • Which of the following is the correct plot for the probability density ψ2 (r) as a function of distance 'r' of the electron from the nucleus for 2s orbital?2022 · MCQ
  • The number of radial nodes and total number of nodes in 4p orbital respectively are :2022 · MCQ
  • A 50 watt bulb emits monochromatic red light of wavelength of 795 nm. The number of photons emitted per second by the bulb is x × 1020. The value of x is ​. [Given : h = 6.63 × 10 − 34 Js and c =…2021 · Numerical
  • When light of wavelength 248 nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electrons is ​Ao​. (Round off to the Nearest Integer). [ Use : 3​= 1.73, h =…2021 · Numerical