Light of wavelength `lambda` falls on metal having work functions `hc//lambda_(0)` . Photoelectric effect will take place only if :
A. `lamdagelamda_(0)`
B. `lamdalelamda_(0)`
C. `lamdage2lamda_(0)`
D. `lamda=4lamda_(0)`
1 Answers
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Light of wavelength 488 nm is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on
Given, `lambda = 488 nm = 488xx10^(-9)m, V_(0)=0.38 V, e=1.6xx10^(-19)c, h=6.63xx10^(-34)J-s` `c=3xx10^(8)m//s rArr KE = eV_(0)=(hc)/(lambda) - phi_(0) rArr 1.6xx10^(-19)xx0.38=(6.63xx10^(-34)xx3xx10^(8))/(488xx10^(-9))- phi_(0)` `6.08xx10^(-20)=40.75xx10^(-20)-phi_(0) rArr phi_(0) = (40.75-6.08)xx10^(-20)=34.67xx0^(-20)J` `= (34.67xx10^(-20))/(1.6xx10^(-19))eV therefore phi_(0)=2.17...
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If `b-c,2b-lambda,b-a " are in HP, then " a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are is
Correct Answer - B `(2b-lambda)=(2(b-c)(b-a))/((b-c)+(b-a))` `implies (2b-lambda)=(2b-(a+c))=2[b^(2)-(a+c)b+ac]` `implies 2b^(2)-2blambda+lambda (a+c)-2ac=0` `implies b^(2)-blambda+(lambda)/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)-lambda^(2)/(4)+lambda/(2)(a+c)-ac=0` `implies (b-(lambda)/(2))^(2)=lambda^(2)/(4)-(lambda)/(2)(a+c)+ac` `implies (b-(lambda)/(2))^(2)=(a-lambda/(2))(c-(lambda)/(2))` Hence, `a-(lambda)/(2),b-(lambda)/(2),c-(lambda)/(2)` are in GP.
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If `sum_(i=1)^(n)a_(i)^(2)=lambda, AAa_(i)ge0` and if greatest and least values of `(sum_(i=1)^(n)a_(i))^(2)` are `lambda_(1)` and `lambda_(2)` respec
Correct Answer - B `:.`AM of 2nd powes `ge` 2nd power of AM `:.(a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2))/(n) ge((a_(1)+a_(2)+a_(3)+"..."a_(n))/(n))^(2)` `implies (lambda)/(n)ge((sum_(i=1)^(n)a_(i))/(n))^(2) :.ge(sum_(i=1)^(n)a_(i))^(2)le n lambda " " ".......(i)"` Also,`(a_(1)+a_(2)+a_(3)+"..."a_(n))^(2)=a_(1)^(2)+a_(2)^(2)+a_(3)^(2)+"..."a_(n)^(2)+2suma_(1)a_(2)=lambda+2suma_(1)a_(2)gelambda` `:.(sum_(i=1)^(n)a_(i))^(2)le lambda" " ".......(ii)"` From Eqs. (i)...
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If the matrix `A = [[lambda_(1)^(2), lambda_(1)lambda_(2), lambda_(1) lambda_(3)],[lambda_(2)lambda_(1),lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(
Correct Answer - 1 `because A^(2) = Acdot A = [[lambda_(1)^(2), lambda_(1) lambda_(2), lambda_(1) lambda_(3) ],[lambda_(2)lambda_(1), lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]] [[lambda_(1)^(2), lambda_(1) lambda_(2), lambda_(1) lambda_(3) ],[lambda_(2)lambda_(1), lambda_(2)^(2),lambda_(2)lambda_(3)],[lambda_(3)lambda_(1),lambda_(3)lambda_(2),lambda_(3)^(2)]]` ` = [[lambda_(1)^(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)), lambda_(1) lambda_(2)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2)), lambda_(1) lambda_(3)(lambda_(1)^(2)+lambda_(2)^(2)+lambda_(3)^(2))...
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A ray of light of wavelength `lambda` is incident on mirror at an angle of incidence `60^(@)`. The wavelength of light after reflection will be
Correct Answer - D Due to reflection there is no change in wavelength, velocity and frequency.
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हाइड्रोजन स्पेक्ट्रम में प्राप्त लाइमन बामर एवं पाश्च्न की सूक्ष्मतम तरंगदैर्ध्य को क्रमशः `lambda_(L), lambda_(B)` एवं `lambda_(P)` से व्यक्त किया जा
`lambda_(L)ltlambda_(B)ltlambda_(P)`
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Assertion : Photoelectric effect supports the quantum nature of light. Reason : Photoelectric emission is instantaneous.
Correct Answer - B (b) : Photoelectric effect is observed only when the frequency value depending upon the metal. Small amount of energy given slowly over a long time does not...
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`lambda_(CICH_(2)COONa)=224ohm^(-1)cm^(2)` gm `eq^(-1)` `lambda_(NaCl)=38.2 ohm^(-1)cm^(2)` gm `eq^(-1) lambda_(HCl)`=203 `ohm^(-1)cm^(2)gm eq^(-1)`,
Correct Answer - C `lambda_(ClCH_(2)COOH)=lambda_(ClCH_(2)COONa)+ lambda_(HCl)-lambda_(NaCl)` `=(224+203-38.2)ohm^(-1)cm^(2)(g eq^(-1))` =`388.8ohm^(-1)cm^(2)(g eq^(-1))`
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The threshold wavelength for the ejection of electrons from the metal X is 330 nm. The work function for the photoelectric emission for metal X is `(h
Correct Answer - B Work function is given as =`hv^@ =(hc)/lambda^@` or Work function =`(6.62xx10^(-34)xx3xx10^8)/(330xx10^(-9))` `=6xx10^(-19) J`
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If `lambda_(Cu)` is the wavelength of `K_(alpha)` X-ray line of copper (atomic number `29`) and `lambda_(Mo)` is the wavelength of the `K_(alpha)`X-ra
Correct Answer - B `sqrt(C/lambda)=a (Z-b)` `b=1 sqrt(lambda_(Cu)/lambda_(Me))=((Z_(Me)-1)/(Z_(Cu)-1))` `lambda_(Cu)/lambda_(Me)=(41/28)^(2)=2.14`
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