Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

The work function of a surface of a photosensitive material  is 6.2 eV. The wavelength of the incident radiation forwhich the stopping potential is 5 eV lies in the

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

ultraviolet region

b

visible region

c

infrared region

d

X-ray region

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

According to laws of photoelectric effectKEmax=E−ϕwhere  ϕ  is  work  function  and  KEmax,  is  maximum,kinetic  energy  of  photoelectron.∴hv=eVo+ϕor   hv=5eV+6.2eV=11.2eV∴λ=1240011.2Ao≈1000AoHence, the radiation lies in ultraviolet region.
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring