Top 50 Most Repeated DUAL NATURE OF RADIATION AND MATTER PYQs | JEE MAINS
A curated collection of the most important questions from DUAL NATURE OF RADIATION AND MATTER, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from DUAL NATURE OF RADIATION AND MATTER, fully solved with step-by-step concepts to prepare for JEE MAINS.
First Lyman line: n=21, second: n=31. Using $\frac{1}{\lambda} \propto \left(1 - \frac{1}{n^2}\right)$, $\lambda_2 = \frac{8}{9} \times 1216 \approx 1...
Read Full Step-by-Step Solution →In $\beta$-decay, a neutron converts to proton, increasing atomic/proton number by 1....
Read Full Step-by-Step Solution →Using the Rydberg formula, \(\frac{1}{\lambda}=R_\infty\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.097\times10^{7}\times\frac{5}{36}=1.525\times10^...
Read Full Step-by-Step Solution →Step 1: Substitute the given expression for $K$ into the equation for $\frac{1}{\lambda}$.\nStep 2: Simplify the resulting expression to obtain the fi...
Read Full Step-by-Step Solution →Cutoff wavelength $\lambda_{\min} = \frac{hc}{eV}$ depends only on accelerating voltage $V$....
Read Full Step-by-Step Solution →Using $\lambda = \frac{1.226}{\sqrt{V}}$ nm, $\lambda = \frac{1.226}{10} = 0.1226 \approx 0.12$ nm....
Read Full Step-by-Step Solution →$\Delta m = (1.00727+1.00866) - 2.01355 = 0.00238u$. $BE = 0.00238 \times 931.5 \approx 2.22$ MeV....
Read Full Step-by-Step Solution →$\Delta m = 0.1\,u$, $E = 0.1 \times 931.5 = 93.15\,\text{MeV}$, per nucleon: $93.15/60 \approx 9.30\,\text{MeV}$...
Read Full Step-by-Step Solution →Alpha decay: $^A_ZX \to ^{A-4}_{Z-2}Y + ^4_2He$. So $Z$ decreases by 2, $A$ by 4....
Read Full Step-by-Step Solution →Threshold frequency $\nu_0 = \frac{\phi}{h} = \frac{2.5}{4.14 \times 10^{-15}} \approx 6.04 \times 10^{14}$ Hz. Rounded to nearest integer in $\times ...
Read Full Step-by-Step Solution →Using $K_{\max} = h\nu - \phi = \frac{1240}{\lambda(\text{nm})} - 4.2$, gives $K_{\max} = 0.7 \text{ eV}$....
Read Full Step-by-Step Solution →Use $\phi = h f_0$. $f_0 = \frac{4.2}{4.14 \times 10^{-15}} \approx 1.01 \times 10^{15}$ Hz →
First Lyman line: n=21, second: n=31. Using $\frac{1}{\lambda} \propto \left(1 - \frac{1}{n^2}\right)$, $\lambda_2 = \frac{8}{9} \times 1216 \approx 1...
Read Full Step-by-Step Solution →In $\beta$-decay, a neutron converts to proton, increasing atomic/proton number by 1....
Read Full Step-by-Step Solution →Using the Rydberg formula, \(\frac{1}{\lambda}=R_\infty\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.097\times10^{7}\times\frac{5}{36}=1.525\times10^...
Read Full Step-by-Step Solution →Step 1: Substitute the given expression for $K$ into the equation for $\frac{1}{\lambda}$.\nStep 2: Simplify the resulting expression to obtain the fi...
Read Full Step-by-Step Solution →Cutoff wavelength $\lambda_{\min} = \frac{hc}{eV}$ depends only on accelerating voltage $V$....
Read Full Step-by-Step Solution →Using $\lambda = \frac{1.226}{\sqrt{V}}$ nm, $\lambda = \frac{1.226}{10} = 0.1226 \approx 0.12$ nm....
Read Full Step-by-Step Solution →$\Delta m = (1.00727+1.00866) - 2.01355 = 0.00238u$. $BE = 0.00238 \times 931.5 \approx 2.22$ MeV....
Read Full Step-by-Step Solution →$\Delta m = 0.1\,u$, $E = 0.1 \times 931.5 = 93.15\,\text{MeV}$, per nucleon: $93.15/60 \approx 9.30\,\text{MeV}$...
Read Full Step-by-Step Solution →Alpha decay: $^A_ZX \to ^{A-4}_{Z-2}Y + ^4_2He$. So $Z$ decreases by 2, $A$ by 4....
Read Full Step-by-Step Solution →Threshold frequency $\nu_0 = \frac{\phi}{h} = \frac{2.5}{4.14 \times 10^{-15}} \approx 6.04 \times 10^{14}$ Hz. Rounded to nearest integer in $\times ...
Read Full Step-by-Step Solution →Using $K_{\max} = h\nu - \phi = \frac{1240}{\lambda(\text{nm})} - 4.2$, gives $K_{\max} = 0.7 \text{ eV}$....
Read Full Step-by-Step Solution →Use $\phi = h f_0$. $f_0 = \frac{4.2}{4.14 \times 10^{-15}} \approx 1.01 \times 10^{15}$ Hz →
First Lyman line: n=21, second: n=31. Using $\frac{1}{\lambda} \propto \left(1 - \frac{1}{n^2}\right)$, $\lambda_2 = \frac{8}{9} \times 1216 \approx 1...
Read Full Step-by-Step Solution →In $\beta$-decay, a neutron converts to proton, increasing atomic/proton number by 1....
Read Full Step-by-Step Solution →Using the Rydberg formula, \(\frac{1}{\lambda}=R_\infty\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.097\times10^{7}\times\frac{5}{36}=1.525\times10^...
Read Full Step-by-Step Solution →Step 1: Substitute the given expression for $K$ into the equation for $\frac{1}{\lambda}$.\nStep 2: Simplify the resulting expression to obtain the fi...
Read Full Step-by-Step Solution →Cutoff wavelength $\lambda_{\min} = \frac{hc}{eV}$ depends only on accelerating voltage $V$....
Read Full Step-by-Step Solution →Using $\lambda = \frac{1.226}{\sqrt{V}}$ nm, $\lambda = \frac{1.226}{10} = 0.1226 \approx 0.12$ nm....
Read Full Step-by-Step Solution →$\Delta m = (1.00727+1.00866) - 2.01355 = 0.00238u$. $BE = 0.00238 \times 931.5 \approx 2.22$ MeV....
Read Full Step-by-Step Solution →$\Delta m = 0.1\,u$, $E = 0.1 \times 931.5 = 93.15\,\text{MeV}$, per nucleon: $93.15/60 \approx 9.30\,\text{MeV}$...
Read Full Step-by-Step Solution →Alpha decay: $^A_ZX \to ^{A-4}_{Z-2}Y + ^4_2He$. So $Z$ decreases by 2, $A$ by 4....
Read Full Step-by-Step Solution →Threshold frequency $\nu_0 = \frac{\phi}{h} = \frac{2.5}{4.14 \times 10^{-15}} \approx 6.04 \times 10^{14}$ Hz. Rounded to nearest integer in $\times ...
Read Full Step-by-Step Solution →Using $K_{\max} = h\nu - \phi = \frac{1240}{\lambda(\text{nm})} - 4.2$, gives $K_{\max} = 0.7 \text{ eV}$....
Read Full Step-by-Step Solution →Use $\phi = h f_0$. $f_0 = \frac{4.2}{4.14 \times 10^{-15}} \approx 1.01 \times 10^{15}$ Hz →
First Lyman line: n=21, second: n=31. Using $\frac{1}{\lambda} \propto \left(1 - \frac{1}{n^2}\right)$, $\lambda_2 = \frac{8}{9} \times 1216 \approx 1...
Read Full Step-by-Step Solution →In $\beta$-decay, a neutron converts to proton, increasing atomic/proton number by 1....
Read Full Step-by-Step Solution →Using the Rydberg formula, \(\frac{1}{\lambda}=R_\infty\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=1.097\times10^{7}\times\frac{5}{36}=1.525\times10^...
Read Full Step-by-Step Solution →Step 1: Substitute the given expression for $K$ into the equation for $\frac{1}{\lambda}$.\nStep 2: Simplify the resulting expression to obtain the fi...
Read Full Step-by-Step Solution →Cutoff wavelength $\lambda_{\min} = \frac{hc}{eV}$ depends only on accelerating voltage $V$....
Read Full Step-by-Step Solution →Using $\lambda = \frac{1.226}{\sqrt{V}}$ nm, $\lambda = \frac{1.226}{10} = 0.1226 \approx 0.12$ nm....
Read Full Step-by-Step Solution →$\Delta m = (1.00727+1.00866) - 2.01355 = 0.00238u$. $BE = 0.00238 \times 931.5 \approx 2.22$ MeV....
Read Full Step-by-Step Solution →$\Delta m = 0.1\,u$, $E = 0.1 \times 931.5 = 93.15\,\text{MeV}$, per nucleon: $93.15/60 \approx 9.30\,\text{MeV}$...
Read Full Step-by-Step Solution →Alpha decay: $^A_ZX \to ^{A-4}_{Z-2}Y + ^4_2He$. So $Z$ decreases by 2, $A$ by 4....
Read Full Step-by-Step Solution →Threshold frequency $\nu_0 = \frac{\phi}{h} = \frac{2.5}{4.14 \times 10^{-15}} \approx 6.04 \times 10^{14}$ Hz. Rounded to nearest integer in $\times ...
Read Full Step-by-Step Solution →