Search Colleges, Courses, Exams, Questions and Articles

Ask a query on 8879100846 Ask Login | Sign Up

 

 
› Tags › Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

4

Active Users

0

Followers

New answer posted

9 months ago

Class 11th Structure of Atoms

Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 1015 (Hz) [1/32 – 1 /n2]. Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
0 Follower 9 Views

A
alok kumar singh

Contributor-Level 10

Given, ν = (3.29 x 1015 Hz) (1/32 – 1/n2)

= (3.29 x 1015 Hz) (1/32 – 1/n2)

(3 x 108 ms-1) / (1.285 x 10-6 m) = (3.29 x 1015 Hz) (1/32 – 1/n2)

2.3346 x 1014 = (3.29 x 1015 Hz) (1/32 – 1/n2)

2.3346 / 32.9 = 1/32 – 1/n2

0.071 = 1/9 – 1/n2

1/n2 = 1/9 – 0.071= 0.111 – 0.071 = 0.04

n2 = 1/ 0.04 = 25

=> n = 5

For n = 5, Paschen series lies in infrared region of the spectrum.

New answer posted

9 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Thirteen

Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.
0 Follower 4 Views

P
Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1-20 KJ mol-1.

New answer posted

9 months ago

Class 11th Structure of Atoms

If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.
0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

λ = 150 pm, v = 1.5 x 107 m s-1

Kinetic energy, K.E. = ½ mv2 = ½ x 9.1 x 10-31 kg x (1.5 x 107 ms-1)2

= [ (9.1 x 1.5 x 1.5) / 2] x 10-31 +14

= 10.2375 x 10-17 J = 1.02375 x 10-16 J

K.E = hc/ λ = (6.626 x 10-34 kg m2 s-1) / (3 x 108 ms-1) / (1.5 x 10-10 m)

= [ (6.626 x 3) x 10-34+8+10] / 1.5  

= 13.252 x 10-16 J

We know, E = W0 + K.E.

W0 = E – K.E.  = (13.252 – 1.024) x 10-16 J

= 12.228 x 10-16 J

= 12.228 10-16 / 1.602 x 10-19

= 7.63 x 103 eV

New answer posted

9 months ago

Class 11th Structure of Atoms

The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
0 Follower 17 Views

A
alok kumar singh

Contributor-Level 10

λ = 256.7 nm = 256.7 x 10-9 m

K.E. = 0.35 eV

E = hc/ λ = (6.626 x 10-34Js) / (3 x 108 ms-1) / (256.7 x 10-9 m)

= (6.626 x 3 x 10-17) J/ 256.7

= (6.626 x 3 x 10-17) / (256.7 x 1.602 x 10-19) eV

E = 4.83 eV

The potential applied to silver gets converted into kinetic energy of the photoelectron.

So, Kinetic energy, K.E= 0.35 V

=> K.E= 0.35 eV          

E = W0 + K.E.

=> W0 = E – K.E.

= 4.83 eV – 0.35 eV = 4.48 eV.

New answer posted

9 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Thirteen

Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?
0 Follower 14 Views

P
Payal Gupta

Contributor-Level 10

This is a short answer type question as classified in NCERT Exemplar

Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.

New answer posted

9 months ago

Class 11th Structure of Atom

Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) Threshold wavelength and, (b) Planck's constant.

λ (nm)      

500

450

400

v * 10–5 (cm s–1)

2.55

4.35

5.35
0 Follower 62 Views

A
alok kumar singh

Contributor-Level 10

Let the threshold wavelength be λ0 nm or λ0 * 10−9 m.

h (ν−ν0) = ½ mv2

hc (1/λ−1/λ0)= ½ mv2

hc [ (1/500*10−9) – (1/ λ0*10−9)] = ½ m (2.55*106)2 . (1)

Similarly,

hc [ (1/450*10−9) – (1/λ0? *10−9? )] = ½? m (4.35*106)2 . (2)

Similarly,

hc [ (1/400*10−9) – (1/λ0? * 10−9? )] = ½? m (5.2 * 106)2  . (3)

Divide equation (2) by (1),

[ (λ0 – 450) /450λ0] x – [500λ0/ (λ0 – 500)] = (4.35/ 2.55)2

(λ0 – 450)/ (λ0 – 500) = 2.61

λ0= 531 nm.

This is the threshold wavelength.

The value of the threshold wavelength is substituted in equation (3).

h *3*108 (1/400*10−9– 1/531*10−9)= ½ *9.1

...more

New answer posted

9 months ago

Class 11th Structure of Atoms

The work function for caesium atom is 1.9 eV. Calculate (a) The threshold wavelength and (b) The threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.       
0 Follower 30 Views

A
alok kumar singh

Contributor-Level 10

W0 = 1.9 eV = 1.9 x 1.602 x 10-19 J

Threshold frequency, v0 = W0 / h = (1.9 x 1.602 x 10-19 J) / (6.626 x 10-34Js)

= 0.459 x 1015 s-1 = 4.59 x 1014 s-1

Threshold wavelength? 0 = c / v0 = (3 x 108 ms-1) / (4.59 x 1014 s-1)

= 0.6536 x 10-6 m = 653.6 nm? 654 nm

Kinetic energy, E = E0 + ½ mv2

(1/2 mv2) = E – E0 = hc [ (1/? ) – (1/? 0)]

= (6.626 x 10-34Js) x (3 x 108 ms-1) / (10-9) x [ (1/500) – (1/654)]

= 6.626 x 3 x 154 x 10-34+8+9) / (500 x 654)

= 9.36 x 10-20 J

Velocity, v = [ (2 x 9.36 x 10-20) / m]1/2

= [ (2 x 9.36 x 10-20) kg m2 s-2 / 9.1 x 10-31 kg]1/2

= (2.057 x 1011 m2s-2)1/2= (20.57 x 1010 m2s-2)1/2

= 4.5356 x 105 ms-1

New answer posted

9 months ago

Class 11th Structure of Atom

The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states
0 Follower 3 Views

A
alok kumar singh

Contributor-Level 10

λ1 = 589 nm = 589 x 10-9 m

ν1 = c / λ1 = (3 x 108 ms-1) / (589 x 10-9 m) = 5.0934 x 1014 s-1

λ2 = 589.6 nm = 589.6 x 10-9 m

ν2 = c / λ2 = (3 x 108 ms-1) / (589.6 x 10-9 m) = 5.0882 x 1014 s-1

ΔE = E1 – E2 = h [ν1 – ν2]

= (6.626 x 10-34Js) x [ (5.0934 x 1014 s-1) – (5.0882 x 1014 s-1)

= 3.31 x 10-22 J

New answer posted

9 months ago

Class 11th Structure of Atoms

Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.
0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

Time duration, t = 2 ns = 2 x 10-9 s

Frequency, ν = 1 / t = 1 / 2 x 10-9 s = 109 / 2 s-1

Energy of one photon, E = hν = 6.626 x 10-34Js) x (109 / 2 s-1) = 3.25 x 10-25 J

No. of photons = 2.5 x 105

Energy of source = 3.3125 x 10-25 J x 2.5 x 1015 = 8.28 x 10-10 J

New answer posted

9 months ago

Class 11th Structure of Atoms

In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10-18 J from the radiations of 600 nm, calculate the number of photons received by the detector.
0 Follower 13 Views

A
alok kumar singh

Contributor-Level 10

Energy of one photon, E = hν = 6.626 x 10-34Js) x (3 x 108 / 2 ms-1) / 600 x 10-9

= 3.31 x 10-19 J

No. of photons = (3.15 x 10-18) / 3.31 x 10-19 = 9.52 ≈ 10.

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 688k Reviews
  • 1850k Answers

Share Your College Life Experience

Write a Review
Community GuidelinesUser Points System
QuestionsDiscussionsTags

×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.